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.

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.

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


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

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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

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√

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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

[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

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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

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

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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]

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]

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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

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

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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

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

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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√

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

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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

               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

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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

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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∘

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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°

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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 :

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

(iii) False Because
             Clearly from the table below :

              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]

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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]

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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°

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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°

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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

⇒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

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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

⇒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√

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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

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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.