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      Class 10 Maths

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      • Class 10
      • Class 10 Maths
      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        NCERT Solutions – Introduction to Trigonometry

        Q.1     In ΔABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine :
                   (i) sin A, cos A               (ii) sin C, cos C
        Sol.       Let us draw a right ΔABC.
        1
        By using the Pythagoras theorem, we have
        AC2=AB2+BC2=(24)2+(7)2
        = 576 + 49 = 625
        ⇒AC=625−−−√=25cm
        (i) sinA=BCAC=725, [Since, sinθ=PH]
        cosA=ABAC=2425     [Since,cosθ=BH]

        (ii) sinC=ABAC=2425,
        cosC=BCAC=725.


        Q.2     In adjoining figure, find tan P – cot R.

        2
        Sol.      By using the Pythagoras theorem, we have
        PR2=PQ2+QR2
        ⇒132=122+QR2
        ⇒ QR2=132−122=169−144=25
        ⇒ QR=25−−√=5
        Therefore, tanP=PB=QRPQ=512 and cotR=BP=QRPQ=512
        Hence, tan P – cot R = 512−512=0


        Q.3     If sinA=34, calculate cos A and tan A.
        Sol.       Consider a ΔABC in which ∠B=90∘.
        For ∠A, we have
        Base = AB, Perp = BC and Hyp = AC
        100
        Therefore, sinA=PH,=BCAC=34
        Let BC = 3k and AC = 4k.
        Then, AB=AC2−BC2−−−−−−−−−−√=(4k)2−(3k)2−−−−−−−−−−−√
        =16k2−9k2−−−−−−−−−√=7k2−−−√=7k−−√
        Therefore, cosA=BH=ABAC=7k√4k=7√4
        tanA=PB=BCAB=3k7k√=37√


        Q.4     Given 15 cot A = 8, find sin A and sec A.
        Sol.       Consider a ΔABC in which ∠B=90∘.
        For ∠A, we have
        Base = AB, Perp = BC
        and Hyp = AC

        104
        [Since, 15 cot A = 8 ⇒cotA=815 ]
        Let AB = 8k and BC = 15k.
        Then, AC=AB2+BC2−−−−−−−−−−√=(8k)2+(15k)2−−−−−−−−−−−−√
        =64k2+225k2−−−−−−−−−−−√=289k2−−−−−√=17k
        Therefore, sinA=PH=BCAC=15k17k=1517
        and, secA=HB=ACAB=17k8k=178


        Q.5     Given secθ=1312, calculate all other trigonometric ratios. 
        Sol.       Consider a ΔABC in which ∠A=θ∘ and ∠B=90∘
                     Then, Base = AB, Perp = BC and Hyp = AC
                     Therefore, secθ=HB=ACAB=1312 
                     Let AC = 13k and AB = 12k. Then,
        BC=AC2−AB2−−−−−−−−−−√=(13k)2−(12k)2−−−−−−−−−−−−√
        =169k2−144k2−−−−−−−−−−−√=25k2−−−−√=5k
        Therefore, sinθ=PH=BCAC

        3
        =5k13k=513
        cosθ=BH=ABAC=12k13k=1213
        tanθ=PB=BCAB=512
        cotθ=1tanθ=125
        cosecθ=1sinθ=135


        Q.6      If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A=∠B.
        Sol.      In rt ΔABC, cosA=ACAB=[BH]

        444
        and, cosB=BCAB
        But, cos A = cos B [Given]
        ⇒ACAB=BCAB
        ⇒AC=BC
        Since, in ΔABC, AC = BC
        ⇒∠A=∠B [Angles opposite to equal sides are equal]


        Q.7     If cotθ=78, evaluate :
                   (i) (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)
                   (ii) cot2θ
        Sol.       Consider a ΔABC in which ∠A=θ∘ and ∠B=90∘. Then, Base = AB, Perp = BC and Hyp = AC
        5
        Therefore, cotθ=BP=ABBC=78
        Let AB = 7k and BC = 8k.
        Then, AC=BC2+AB2−−−−−−−−−−√=(8k)2+(7k)2−−−−−−−−−−−√
        =64k2+49k2−−−−−−−−−−√=113k2−−−−−√=113k−−−−√
        Therefore, sinθ=PH=BCAC=8k113k√=8113√
        and cosθ=BH=ABAC=7k113k√=7113√

        (i) Therefore, (1+sinθ)(1−sinθ)(1+cosθ)(1−cosθ)=1−sin2θ1−cos2θ=1−641131−49113
        =113−64113−49=4964

        (ii) cot2θ=(78)2=4964


        Q.8     If 3 cot A = 4, check whether 1−tan2A1+tan2A=cos2 A−sin2A or not.
        Sol.      Consider a ΔABC in which ∠B=90∘.
        For ∠A, we have
        Base = AB, Perp = BC and Hyp = AC

        6
        Therefore, cotA=BP=ABBC=43       
                     Let AB = 4k and BC = 3k [3cot A = 4 ⇒cotA=43]
        Then, AC=AB2+BC2−−−−−−−−−−√=(4k)2+(3k)2−−−−−−−−−−−√
        =16k2+9k2−−−−−−−−−√=25k2−−−−√=5k
        Therefore, sinA=PH=BCAC=3k5k=35
        cosA=BH=ABAC=4k5k=45
        and, tanA=1cotA=34
        L.H.S. =1−tan2A1+tan2A=1−9161+916=16−916+9=725
        R.H.S. = cos2A−sin2A
        =(45)2−(35)2=1625−925=725
        ⇒ L.H.S = R.H.S.
        Therefore, 1−tan2A1+tan2A=cos2A−sin2A


        Q.9     In ΔABC right angled at B, if tan A = 13√, find the value of
                   (i) sin A cos C + cos A sin C
                   (ii) cos A cos C – sin A sin C
        Sol.       Consider a ΔABC , in which ∠B=90∘.
        For ∠A, we have
        Base = AB, Perp = BC
        and Hyp = AC
        tanA=PerpHyp
        = BCAB=13√
        7
        Let BC = k and AB = 3–√ k .
        Then, AC=AB2+BC2−−−−−−−−−−√=3k2+k2−−−−−−−√
        = 4k2−−−√=2k
        Therefore, sinA=PerpHyp=BCAC=k2k=12
        cosA=BaseHyp=ABAC=3k√2k=3√2
        For ∠C, we have
        Base = BC, Perp = AB and Hyp = AC
        Therefore, sinC=PerpHyp=ABAC=3k√2k=3√2
        and, cosC=BaseHyp=BCAC=k2k=12

        (i) sinA cosC + cosA sinC = 12×12+3√2×3√2
        =14+34=44=1

        (ii) cos A cosC – sin A sin C = 3√2×12−12×3√2=0


        Q.10     In ΔPQR, right angled at Q, PR + QR 25 cm and PQ = 5 cm. Determine the values of sin P cos P and tan P.
        Sol.         In ΔPQR, right ∠d at Q,
        PR + QR = 25 cm and PQ = 5cm
        Let QR = x cm

        105               Therefore, PR = (25 – x) cm
        By Pythagoras theorem, we have
        RP2=RQ2+QP2
        ⇒ (25−x)2=x2+52
        ⇒625−50x+x2=x2+25
        ⇒−50x=−600
        ⇒x=−600−50=12
        Therefore, RQ = 12cm
        ⇒ RP = (25 – 12)cm = 13cm
        Now, sinP=RQRP=1213
        cosP=PQRP=513
        and tanP=RQPQ=125


        Q.11     State whether the following are true or false. Justify your answer.
                     (i) The value of tan A is always less than 1.
                     (ii) secA=125 for some value of angle A .
                     (iii) cos A is the abbreviation used for the cosecant of angle A.
                     (iv) cot A is the product of cot and A.
                     (v) sinθ=43 for some angle θ.
        Sol.          (i) False because sides of a right triangle may have any length, so tan A may have any value.
        (ii) True as sec A is always greater than 1.
        (iii) False as cos A is the abbreviation used for cosine A.
        (iv) False as cot A is not the product of ‘cot’ and A. ‘cot’ separated from A has no meaning.
        (v)  False as sinθ cannot be > 1.

         

        Q.1     Evaluate :
                   (i) sin 60 ° cos 30° + sin 30° cos 60°
                   (ii) 2tan245∘+cos230∘−sin260∘ 
                   (iii) cos45∘sec30∘+cosec30∘
                   (iv) sin30∘+tan45∘−cosec60∘sec30∘+cos60∘+cot45∘
                    (v) 5cos260∘+4sec230∘−tan245∘sin230∘+cos230∘
        Sol.     (i) sin 60° cos 30° + sin 30° cos 60°
                    =3√2×3√2+12×12
        =34+14=3+14=44=1

        (ii) 2tan245∘+cos230∘−sin260∘
        =2(1)2+(3√2)2−(3√2)2
        =2+34−34=2

        (iii) cos45∘sin30∘+cosec30∘
        =12√23√+2=12√2+23√3√=12√×3√2+23√
        = 3√2√×2(3√+1)×3√−13√−1
        = 3√(3√−1)2√×2(3−1)=2√×3√(3√−1)2√×2√×2×2
        = 32√−6√8

        (iv) sin30∘+tan45∘−cosec60∘sin30∘+cos45∘+cot45∘
                     12+1−23√23√+12=1+22−23√23√+1+22
        =32−23√23√+32=33√−423√4+33√23√
        =33√−44+33√=(33√−4)(33√−4)(4+33√)(33√−4)
        27+16−123√−123√27−16
        43−243√11

        (v) 5cos260∘+4sec230∘−tan245∘sin230∘+cos230∘
        Sol.       5(12)2+4(23√)2−(1)2(12)2+(3√2)2
        5×14+4×43−114+34=54+163−1=112(15+64−12)1+34
        112×6744=6712


        Q.2     Choose the correct option and justify :
                   (i) 2tan30∘1+tan230∘=
                   (A) sin 60°                    (B) cos 60°
                   (C) tan 60°                   (D) sin 30°

                   (ii) 1−tan245∘1+tan245∘=
                   (A) tan 90°                   (B) 1
                   (C) sin 45 °                   (D) 0

                   (iii) sin 2A = 2 sinA is true when A =
                    (A) 0°                           (B) 30°
                    (C) 45°                         (D) 60°

                    (iv) 2tan230∘1−tan230∘=
                    (A) cos 60°                   (B) sin 60°
                    (C) tan 60 °                  (D) none of these
        Sol.        (i) (A)
        Because 2tan30∘1+tan230∘=2×13√1+(13√)2=23√1+13
                       =23√×33+1=23√×34
                       =3√2=sin60∘

        (ii) (D)
        Because 1−tan245∘1+tan245∘=1−11+1=02=0

        (iii) (A)
        Because when A = 0, sin 2 A = sin 0 = 0
        and, 2 sinA = 2 sin 0 = 2 × 0 = 0
        ⇒ sin 2A = 2sinA, when A = 0

        (iv) (C)
        Because 2tan30∘1−tan230∘=2×13√1−(13√)2=23√1−13
        =23√×33−1=23√×32
        =3–√=tan60∘


        Q.3     If tan (A + B) = 3–√ and tan (A – B) = 13√;0∘<A+B≤90∘;A>B, find A and B.
        Sol.       tan (A + B) = 3–√
        ⇒tan(A+B)=tan60∘
        ⇒A+B=60∘ …(1)
        tan(A−B)=13√
        ⇒tan(A−B)=tan30∘
        ⇒ A – B = 30 °…(2)
        Solving (1) and (2), we get
        A = 45° and B = 15°
        Hence, A = 45° and B = 15°


        Q.4     State whether the following are true of false. Justify your answer.
                   (i) sin (A + B) = sin A + sin B.
                   (ii) The value of sin θ increases as θ increases.
                   (iii) The value of casθ increases as θ increases.
                   (iv) sinθ = cosθ for all values of θ.
                   (v) cot A is not defined for A = 0°.
        Sol.       (i) False Because
        When A = 60° and B = 30°. Then,
        sin (A + B) = sin (60° + 30°) = sin 90° = 1
        and, sin A + sin B = sin 60° + sin 30°
        =3√2+12=3√+12
        ⇒ sin (A + B) ≠ sin A + sin B
        (ii) True Because
        Clearly from the table below :

        10
        We find that the value of sinθ increases as θ increases.

        (iii) False Because
                     Clearly from the table below :

        11

                      We find that the value of cosθ decreases as θ increases.

                      (iv)  False as it is only true for θ = 45°.
        (sin45∘=12√=cos45∘)

        (v) True Because tan 0° = 0 and cot0∘=1tan0∘=10,i.e., undefined.

         

        Q.1     Evaluate :
                   (i) sin18∘cos72∘
                   (ii) tan26∘cot64∘
                   (iii) cos 48° – sin 42°
                   (iv) cosec 31° – sec 59°
        Sol.       (i) sin18∘cos72∘=sin(90∘−72)cos72∘=cos72∘cos72∘=1[Since,sin90−θ=cosθ]

        (ii) tan26∘cot64∘=tan(90∘−64∘)cot64∘=cot64∘cot64∘=1 [Since,tan(90−θ)=cotθ]

        (iii) cos48° – sin42° = cos (90° – 42°) – sin 42°
        = sin 42° – sin 42° = 0 [Since,cos(90−θ)=sinθ]
         (iv) cosec 31° – sec 59° = cosec (90° – 59°) – sec 59° = 0
                                                          =sec59∘−sec59∘=0 [Since,cosec(60−θ)=secθ]


        Q.2     Show that
                   (i) tan 48° tan 23° tan 42° tan 67° = 1
                   (ii) cos 38° cos 52° – sin 38° sin 52 ° = 0
        Sol.       (i) tan 48° tan 23° tan 42 ° tan 67°
        = tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°
        = cot 42° cot 67° tan 42° tan 67°  [Since,tan(90−θ)=cotθ]
        =1tan42∘.1tan67∘.tan42∘tan67∘=1

        (ii) cos 38°cos 52°– sin 38°sin 52°
        = cos(90° – 52°)cos(90°– 38°) – sin 38°sin 52°
        = sin 52° sin 38° – sin 38°sin 52° = 0 [Since,cos(90∘−θ)=sinθ]


        Q.3     If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
        Sol.       We are given that
        tan 2A = cot (A – 18°)…(1)
                     Since tan 2A = cot (90° – 2A), so we can write (1) as
        cot(90°–2A) = cot (90° – 2A), so we can write (1) as
        cot (90° – 2A) = cot (A – 18°) [Since,cot(90∘−θ)=tanθ]
        Since (90° – 2A) and (A – 18°) are both acute angle therefore,
        90° – 2A = A – 18°
        ⇒ – 2A – A = – 18° – 90°
        ⇒ – 3A = – 108°
        ⇒ A = 36°


        Q.4     If tan A = cot B, prove that A + B = 90°.
        Sol.       We are given that
        tan A = cot B…(1)
        Since tan A = cot(90° – A), so we can write (1) as
        cot (90° – A ) = cot B [Since,cot(90∘−θ)=tanθ]
        Since (90° – A) and B are both acute angles, therefore,
                      (90° – A) = B
        ⇒ A + B = 90°


        Q.5     If sec 4 A = cosec (A – 20°), where 4 A is an acute angle, find the value of A.
        Sol.      We are given that sec 4A = cosec (A – 20°). ….(1)
        Since sec 4A = cosec (90°– 4A), so we can write (1) as
        cosec (90° – 4A) = cosec (A – 20°) [Since,cosec(90∘−θ)=secθ]
        Since (90° – 4A) and (A – 20°) are both acute angles, therefore,
        90° – 4A = A – 20°
        ⇒ – 4A – A = – 20° – 90°
        ⇒ – 5A = 110°
        ⇒ A = 22°

         

        Q.1     Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
        Sol.       Consider a ΔABC, in which ∠B=90∘.
        For ∠A, we have
        Base = AB, Perp = BC
        and Hyp = AC
        Therefore, cotA=BasePerp=ABBC

        8
        ⇒ABBC=cotA=cotA1
        Let AB = k cotA and BC = k.
        By Pythagoras Theorem,
        AC=AB2+BC2−−−−−−−−−−√=k2cot2A+k2−−−−−−−−−−−√
        =k1+cot2A−−−−−−−−√
        Therefore, sinA=PerpBase=BCAC=kk1+cot2A√
        =11+cot2A√
        secA=HypBase=ACAB=k1+cot2A√kcotA
                                =1+cot2AcotA−−−−−−√
        and, tanA=PerpBase=BCAB=kkcotA=1cotA


        Q.2     Write the other trigonometric ratios of A in terms of secA.
        Sol.        Consider a ΔABC, in which ∠B=90∘.
        For ∠A, we have
        Base = AB, Perp = BC
        and Hyp = AC
        Therefore, secA=HypBase=ACAB
        ⇒ACAB=secA=secA1

        9

        ⇒ABAC=1secA=1secA1 (Note this step)
        Let AB=k(1secA),AC=k(1)
        By phythagoras Theorem
        BC=AC2+AB2−−−−−−−−−−√=k2−k2(1sec2A)−−−−−−−−−−−−√
                             =ksec2A−1sec2A−−−−−−√=ksec2A−1√sec2A
        Therefore, sinA=BCAC=ksec2A−1√secAk=sec2A−1√secA
        cosA=ABACk(1secA)k=1secA
        tanA=BCAB=ksec2A−1√secAk(1secA)=sec2A−1−−−−−−−−√
        cotA=1tanA=1sec2A−1√
        cosecA=1sinA=secAsec2A−1√


        Q.3     Evaluate :
                      (i) sin263∘+sin227∘cos217∘+cos273∘
                    (ii) sin 25° cos 65° + cos25° sin65° [Since,sin(90∘−θ)=cosθ]
         Sol.       (i) Here, sin63° = sin(90° – 27°) = cos27°
                       and cos17° = cos(90° – 73°) = sin 73° [Since,cos(90∘−θ)=sinθ]
                       Therefore, sin263∘+sin227∘cos217∘+cos273∘=cos227∘+sin227∘sin273∘+cos273∘
                        =11=1
        [Since,cos2A+sin2A=1]

        (ii) sin25°cos65°+cos25°sin65°
        = sin(90°– 65°). cos65° + cos(90°– 65°) sin65°
        = cos65° cos65° + sin 65° sin65°
        [Since,sin(90∘−θ)=cosθ]
        = cos265∘+sin265∘=1 [Since,cos(90∘−θ)=sinθ]

        Q.4     Choose the correct option. Justify your choice :
        (i) 9sec2A−9tan2A=
                   (A) 1                                     (B) 9
                   (C) 8                                      (D) 0

                   (ii) (1+  tanθ + secθ)(1 + cos θ – cosec θ) =
                    (A) 0                                     (B) 1
                    (C) 2                                     (D) None of these

                    (iii) (secA + tanA)(1 – sinA) =
                    (A) secA                              (B) sinA
                    (C) cosecA                           (D) cosA

                    (iv) 1+tan2A1+cot2A=
                    (A) sec2A                              (B) –1
                    (C) cot2A                               (D) none of these
        Sol.       (i) (B), because
        9sec2A−9tan2A=9(sec2A−tan2A)
        = 9 × 1 = 9 [Since,1+tan2A=sec2A]

        (ii) (C), because
        (1+tanθ+secθ)(1+cotθ−cosecθ)
        =(1+sinθcosθ+1cosθ)(1+cosθsinθ−1sinθ)
        =(cosθ+sinθ+1cosθ)(sinθ+cosθ−1sinθ)
        = (cosθ+sinθ)2−1sinθcosθ [Since,(A+B)(A−B)=A2−B2]
        =  (cos2θ+sin2θ)−2cosθsinθ−1sinθcosθ [Since,sin2θ+cos2θ=1]
        = 1+2cosθsinθ−1sinθcosθ=2cosθsinθsinθcosθ=2

        (iii) (D), because
        (secA + tanA) (1 – sinA)
        = (1cosA+sinAcosA)(1−sinA)
        = (1+sinAcosA)(1−sinA) [Since,(A+B)(A−B)=A2−B2]
        =1−sin2AcosAcos2AcosA=cosA[Since,sin2A+cos2A=1]

        (iv) (D), because
        1+tan2A1−cot2A=1+tan2A1+1tan2A=1+tan2Atan2A+1tan2A
        =(1+tan2A)×tan2A1+tan2A=tan2A


        Q.5     Prove the following identities, where the angles involved are acute angles for which the expressions are defined :
                   (i) (cosecθ−cotθ)2=1−cosθ1+cosθ
                   (ii) cosA1+sinA+1+sinAcosA=2secA
                   (iii) tanθ1−cotθ+cotθ1−tanθ=1+secθcosecθ
                   (iv) 1+secAsecA=sin2A1−cosA
                   (v) cosA−sinA+1cosA+sinA−1=cosecA+cotA, using the identity cosec2A=1+cot2A
                   (vi) 1+sinA1−sinA=−−−−−−−√secA+tanA
                   (vii) sinθ−2sin3θ2cos3θ−cosθ=tanθ
                   (viii) (sinA+cosecA)2+(cosA+secA)2=7+tan2A+cot2A
                   (ix) (cosecA – sinA)(secA – cosA) = 1tanA+cotA
                   (x) (1+tan2A1+cot2A)=(1−tanA1−cot2A)2=tan2A
        Sol.       (i) We have,
        L.H.S. = (cosecθ−cotθ)2
        =(1sinθ−cosθsinθ)2=(1−cosθsinθ)2
        =(1−cosθ)2sin2θ=(1−cosθ)21−cos2θ [Since,sin2θ=1−cos2θ]
        =(1−cosθ)2(1−cosθ)(1+cosθ)=1−cosθ1+cosθ
        R.H.S. [Since,A2−B2=(A+B)(A−B)]

        (ii) We have,
        L.H.S. = cosA1+sinA+1+sinAcosA
        = cos2A+(1+sinA)2cosA(1+sinA)
        = cos2A+1+2sinA+sin2AcosA(1+sinA)
        = (cos2A+sin2A)+1+2sinAcosA(1+sinA)
        = 1+1+2sinAcosA(1+sinA)[Sincesin2A+cos2A=1]
        = 2+2sinAcosA(1+sinA)=2(1+sinA)cosA(1+sinA)
        = 2cosA=2secA=R.H.S.

        (iii) We have,
        L.H.S. = tanθ1−cotθ+cotθ1−tanθ
        = tanθ1−1tanθ+1tanθ1−tanθ
        =tanθtanθ−1tanθ+1tanθ(1−tanθ)
                       =tan2θtanθ−1+1tanθ(1−tanθ)
        =tan2θtanθ−1−1tanθ(tanθ−1)
        = tan3θ−1tanθ(tanθ−1)
        = (tanθ−1)(tan2θ+tanθ+1)tanθ(tanθ−1)
        [Since,a3−b3=(a−b)(a2+ab+b2)]
        = tan2θ+tanθ+1tan2θ
        = tan2θtanθ+tanθtanθ+1tanθ
        = tanθ+1+cotθ=1+tanθ+cotθ
        = 1+sinθcosθ+cosθsinθ
        = 1+sin2θ+cos2θcosθ
        = 1+1sinθcosθ=1+cosecθsecθ
        = R.H.S.

        (iv) R.H.S. = sin2A1−cosA=1−cos2A1−cosA
                         [Since,sin2A=1−cos2A]
        = (1−cosA)(1+cosA)1−cosA=1+cosA
        [Since,A2−B2=(A+B)(A−B)]
        = 1+1secA=1+secAsecA=L.H.S.

        (v) L.H.S. = cosA−sinA+1cosA+sinA−1=cosA−sinA+1sinAcosA+sinA−1sinA
        = cosA−1+cosecAcosA+1−cosecA
        [Since,1+cot2A=cosec2A]
        =cotA+cosecA−(cosec2A−cot2A)cotA−cosecA+1
        = cotA+cosecA−(cosecA+cotA)(cosecA−cotA)cotA−cosecA+1
        [SinceA2−B2=(A+B)(A−B)]
        Taking common(cosecA + cotA)
        = (cosecA+cotA)(1−cosecA+cot)(cotA−cosecA+1)
        = cosec A + cot A
        = R.H.S.

        (vi) We have,
        L.H.S. = 1+sinA1−sinA−−−−−√=1+sinA1−sinA×1+sinA1+sinA−−−−−−−−−−−−−√
        [Multiplying and dividing by ] 1+sinA−−−−−−−√
        = (1+sinA)21−sin2A−−−−−−−√=(1+sinA)2cos2A−−−−−−−√ [Since,sin2A+cos2A=1]
        = (1+sinAcosA)−−−−−−−−√2=1+sinAcosA
        = 1cosA+sinAcosA=secA+tanA
        = R.H.S. [Since,tanA=sinAcosA]

        (vii) We, have,
        L.H.S. = sinθ−2sin3θ2cos3θ−cosθ=sinθ(1−2sin2θ)cosθ(2cos2θ−1)
        = tanθ[1−2sin2θ2(1−sin2θ)−1]
        = tanθ[1−2sin2θ2−2sin2θ−1]
        = tanθ[1−2sin2θ1−2sin2θ]=tanθ×1
        = tanθ = R.H.S.

                             (viii) We have,
        L.H.S. = (sinA+cosecA)2+(cosA+secA)2
        = (sin2A+cosec2A+2sinAcosecA)+(cos2A+sec2A+2cosAsecA)
        = (sin2A+cosec2A+2sinA.1sinA)+(cos2A+sec2A+2cosA.1cosA)
        = (sin2A+cosec2A+2)+(cos2A+sec2A+2)
        = sin2A+cos2A+cosec2A+sec2A+4 [Since,sin2A+cos2A=1]
        = 1+(1+cot2A)+(1+tan2A)+4
        = 7+tan2A+cot2A
        = [Since,cosec2A=1+cot2Aandsec2A=1+tan2A]
        = R.H.S.

        (ix) We have,
        L.H.S. = (cosec A – sinA) (secA – cosA)
        = (1sinA−sinA)(1cosA−cosA)
        = (1−sin2AsinA)(1−cos2AcosA)
        = cos2AsinA×sin2AcosA
        = sinA cosA
        = sinAcosAsin2A+cos2A [Since,sin2A+cos2A=1]
        Dividing Numerator and Denominator by sinA cosA
        sinAcosAsinAcosAsin2AsinAcosA+cos2AsinAcosA
        = 1sinAcosA+cosAsinA
        = 1tanA+cotA = R.H.S.

        (x) We have,
        L.H.S. = (1+tan2A1+cot2A)=sec2Acosec2A
        = 1cos2A×sin2A1=tan2A
        R.H.S. = (1−tanA1−cotA)2=(1−tanA1−1tanA)2
        = (1−tanAtanA−1tanA)2=(−tanA)2=tan2A
        Therefore, L.H.S. = R.H.S.

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          6 Comments

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          May 28, 2021

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          September 15, 2021

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          November 12, 2021

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