NCERT solution { Trigonometry - III }
MATHEMATICS FOR CLASS - X
* Trigonometry [ PART - III ]
EXERCISE 8.4
1. Express the trigonometric ratios sinA , secA and tanA in terms of cotA.
Ans: 2 2
cosec A = 1 + cot A
1 2
⇒ ––––––––––– = 1 + cot A
2
sin A
2 1
⇒ sin A = –––––––––––
2
1 + cot A
1
⇒ sin A = –––––––––––
2
√ ( 1 + cot A )
2 2
sec A = 1 + tan A
2 1
⇒ sec A = 1 + –––––––––
2
cot A
2
2 cot A + 1
⇒ sec A = –––––––––––
2
cot A
2
√ ( cot A + 1 )
⇒ sec A = –––––––––––––––––
2
√ ( cot A )
2
√ ( cot A + 1 )
⇒ sec A = ––––––––––––––––
cot A
And ,
1
tanA = –––––––
cotA
2. Write all the other trigonometric ratios of ∠A in terms of secA.
Ans: 2 2
sin A + cos A = 1
2 2
⇒ sin A = 1 - cos A
2 1
⇒ sin A = 1 - –––––––
2
sec A
1
⇒ sin A = √ ( 1 - ––––––– )
2
sec A
2
sec A - 1
⇒ sin A = √ ( –––––––––––––)
2
sec A
2
√ ( sec A - 1 )
⇒ sin A = ––––––––––––– ... (i)
sec A
1
cos A = ––––––– . . . (ii)
sec A
2 2
sec A = 1 + tan A
2 2
⇒ tan A = sec A - 1
2
⇒ tan A = √ ( sec A - 1 ) . . . (iii)
2 2
sin A + cos A = 1
2 2
⇒ sin A = 1 - cos A
1 1
⇒ ––––––––– = 1 - –––––––
2 2
cosec A sec A
2
1 sec A - 1
⇒ ––––––––– = –––––––––––––
2 2
cosec A sec A
2
2 sec A
⇒ cosec A = –––––––––––
2
sec A - 1
therefore
sec A
cosec A = ––––––––––––– . . . (iv)
2
√(sec A - 1 )
2 2
sec A = 1 + tan A
2 2
⇒ tan A = sec A - 1
1 2
⇒ ––––––––– = sec A - 1
2
cot A
2 1
⇒ cot A = –––––––––––––
2
√ ( sec A - 1 )
1
⇒ cot A = ––––––––––––– . . . . (v)
2
√ ( sec A - 1 )
3. Evaluate :
2 2
sin 63° + sin 27°
Ans : 2 2
sin (90 - 63°) + sin 27°
2 2
cos 17° + sin 27°
⇒ 1 Ans . . . .
(ii) sin 25° cos 65° + cos 25° sin 65°
Ans :
sin 25° cos( 90 - 25) + cos 25° sin ( 90 - 25)
⇒ sin 25° . sin 25° + cos 25° . cos 25°
2 2
⇒ sin 25° + cos 25°
⇒ 1 Ans . . . .
4. Choose the correct option . Justify your choice.
2 2
(i) 9 sec A - 9 tan A =
(A) 1 (B) 9
(C) 8 (D) 0
Ans : (B)
(ii) ( 1 + tanθ + secθ ) ( 1 + cotθ - cosecθ ) =
(A) 0 (B) 1
(C) 2 (D) -1
Ans : (C)
(iii) (secA + tanA ) ( 1 - sinA ) =
(A) sec A (B) sin A
(C) cosec A (D) cos A
Ans : (D)
2
1 + tan A
(iv) –––––––––––– =
2
1 + cot A
2
(A) sec A (B) -1
2 2
(C) cot A (D) tan A
Ans : (D)
5. Prove the following identities , where the angles involved are acute angles for which for which the expressions are defined.
2 1 - cos θ
(i) ( cosec θ - cot θ ) = ––––––––––
1 + cos θ
Ans :
2
L.H.S = ( cosecθ - cotθ )
2 2
= cosec θ + cot θ - 2 cosec θ cot θ
2
1 cos θ 1 cos θ
= ——— + ———— - 2 ——— . ———
2 2 sin θ sin θ
sin θ sin θ
2
1 cos θ 2 cos θ
= ——— + ———— - —————
2 2 2
sin θ sin θ sin θ
2
1 + cos θ - 2 cos θ
= ————————————
2
sin θ
2
( 1 - cos θ)
= ————————————
2
1 - cos θ
2
( 1 - cos θ)
= ———————————
( 1 - cos θ ) ( 1 + cos θ)
1 - cos θ
= ————— = R.H.S proved . . .
1 + cos θ
cos A 1 + sin A
(ii) ––––––––– + ––––––––––– = 2 sec A
1 + sin A cos A
Ans :
cos A 1 + sin A
L.H.S = ––––––––– + –––––––––––
1 + sin A cos A
2 2
cos A + ( 1 + sin A )
= ––––––––––––––––––––––
( 1 + sin A ) cos A
2 2
cos A + 1 + sin A + 2 sinA
= ––––––––––––––––––––––––––
( 1 + sin A ) cos A
1 + 1 + 2 sinA
= –––––––––––––––––––
( 1 + sin A ) cos A
2+ 2 sinA
= –––––––––––––––––––
( 1 + sin A ) cos A
2 ( 1 + sinA )
= –––––––––––––––––––
( 1 + sin A ) cos A
2
= –––––––––
cos A
= 2 sec A = R.H.S proved . . .
tan θ cot θ
(iii) –––––––––– + –––––––––– = 1 + sec θ cosec θ
1 - cot θ 1 - tan θ
Ans :
tan θ cot θ
L.H.S = ––––––––– + –––––––––
1 - cot θ 1 - tan θ
1
tan θ –––––––
tan θ
= ––––––––––– + ––––––––––––––
1 1 - tan θ
1 - –––––
tan θ
tan θ 1
= ––––––––––– + ––––––––––––––
tan θ - 1 tan θ ( 1 - tan θ )
––––––––––
tan θ
2
tan θ 1
= ––––––––––– + ––––––––––––––
tan θ - 1 tan θ ( 1 - tan θ )
2
tan θ 1
= ––––––––––– + –––––––––––––––––
tan θ - 1 tan θ { - ( tan θ - 1 )}
2
sec θ + tan θ
= –––––––––––––––––
tan θ
2
sec θ tan θ
= ––––––– + –––––––––
tan θ tan θ
2
cos θ
= –––––––––––– + 1
sin θ
cos θ
= –––––––––––––– + 1
2
cos θ x sin θ
1
= ––––––––––––– + 1
sin θ cos θ
= 1 + cosec θ sec θ = R.H.S proved . . .
2
1 + secA sin A
(iv) ––––––––– = ––––––––––
secA 1 - cosA
Ans :
1 + secA
L.H.S = –––––––––
secA
1
1 + –––––––
cos A
= ––––––––––––––––––
1
–––––––––
cos A
cos A + 1
–––––––––––
cos A
= ––––––––––––––––––
1
–––––––––
cos A
( cos A + 1 ) cos A
= ––––––––––––––––––
cos A
= ( cos A + 1 )
1 - cos A
= ( 1 + cos A ) X –––––––––––––––
1 - cos A
2
1 - cos A
= –––––––––––––
1 - cos A
2
sin A
= ––––––––– = R.H.S proved . . .
1 - cos A
cosA - sinA + 1
(v) –––––––––––––––– = cosecA + cotA
cosA + sinA - 1
Ans :
cosA - sinA + 1
L.H.S = ––––––––––––––––
cosA + sinA - 1
cosA - ( sinA - 1 )
= ––––––––––––––––
cosA + ( sinA - 1 )
On rationalizing the denominator
cosA - ( sinA - 1 ) cosA - ( sinA - 1 )
= –––––––––––––––– X ––––––––––––––––
cosA + ( sinA - 1 ) cosA - ( sinA - 1 )
2
{ cosA - ( sinA - 1 )}
= –––––––––––––––––––––––
2 2
{ cosA } - { ( sinA - 1 ) }
2 2
cos A + ( sin A - 1 ) - 2 cos A ( sin A - 1 )
= –––––––––––––––––––––––––––––––––––––
2 2
{ cosA } - { ( sinA - 1 ) }
2 2
cos A + sin A + 1 - 2 sin A - 2 cos A sin A + 2 cos A
= ––––––––––––––––––––––––––––––––––––––––––––––
2 2
cos A - { sin A + 1 - 2 sin A }
1 + 1 - 2 sin A - 2 cos A sin A + 2 cos A
= –––––––––––––––––––––––––––––––––––––
2 2
cos A - sin A - 1 + 2 sin A
2 - 2 sin A - 2 cos A sin A + 2 cos A
= –––––––––––––––––––––––––––––––––––
2 2 2 2
cos A - sin A - ( cos A + sin A ) + 2 sin A
2 + 2 cos A - 2 sin A - 2 cos A sin A
= –––––––––––––––––––––––––––––––––––
2 2 2 2
cos A - sin A - cos A - sin A + 2 sin A
2 ( 1 + cos A ) - 2 sin A ( 1 + cos A )
= –––––––––––––––––––––––––––––––––––
2
- 2 sin A + 2 sin A
2 ( 1 + cos A ) ( 1 - sin A )
= –––––––––––––––––––––––––––
2 sin A (1 - sin A )
1 + cos A
= ––––––––––––
sin A
1 cos A
= –––––– + ––––––
sin A sin A
= cosec A + cot A = R.H.S proved . . .
Ans:
NCERT solution
Recomended for CBSE and BSEB students
* Trigonometry [ PART - III ]
EXERCISE 8.4
1. Express the trigonometric ratios sinA , secA and tanA in terms of cotA.
Ans: 2 2
cosec A = 1 + cot A
1 2
⇒ ––––––––––– = 1 + cot A
2
sin A
2 1
⇒ sin A = –––––––––––
2
1 + cot A
1
⇒ sin A = –––––––––––
2
√ ( 1 + cot A )
2 2
sec A = 1 + tan A
2 1
⇒ sec A = 1 + –––––––––
2
cot A
2
2 cot A + 1
⇒ sec A = –––––––––––
2
cot A
2
√ ( cot A + 1 )
⇒ sec A = –––––––––––––––––
2
√ ( cot A )
2
√ ( cot A + 1 )
⇒ sec A = ––––––––––––––––
cot A
And ,
1
tanA = –––––––
cotA
2. Write all the other trigonometric ratios of ∠A in terms of secA.
Ans: 2 2
sin A + cos A = 1
2 2
⇒ sin A = 1 - cos A
2 1
⇒ sin A = 1 - –––––––
2
sec A
1
⇒ sin A = √ ( 1 - ––––––– )
2
sec A
2
sec A - 1
⇒ sin A = √ ( –––––––––––––)
2
sec A
2
√ ( sec A - 1 )
⇒ sin A = ––––––––––––– ... (i)
sec A
1
cos A = ––––––– . . . (ii)
sec A
2 2
sec A = 1 + tan A
2 2
⇒ tan A = sec A - 1
2
⇒ tan A = √ ( sec A - 1 ) . . . (iii)
2 2
sin A + cos A = 1
2 2
⇒ sin A = 1 - cos A
1 1
⇒ ––––––––– = 1 - –––––––
2 2
cosec A sec A
2
1 sec A - 1
⇒ ––––––––– = –––––––––––––
2 2
cosec A sec A
2
2 sec A
⇒ cosec A = –––––––––––
2
sec A - 1
therefore
sec A
cosec A = ––––––––––––– . . . (iv)
2
√(sec A - 1 )
2 2
sec A = 1 + tan A
2 2
⇒ tan A = sec A - 1
1 2
⇒ ––––––––– = sec A - 1
2
cot A
2 1
⇒ cot A = –––––––––––––
2
√ ( sec A - 1 )
1
⇒ cot A = ––––––––––––– . . . . (v)
2
√ ( sec A - 1 )
3. Evaluate :
2 2
sin 63° + sin 27°
(i) ––––––––––––––––––––
2 2
cos 17° + cos 73°
sin (90 - 63°) + sin 27°
––––––––––––––––––––
2 2
cos 17° + cos (90 - 73°)
2 2
cos 17° + sin 27°
⇒ ––––––––––––––––––––
2 2
cos 17° + sin 27°
⇒ 1 Ans . . . .
(ii) sin 25° cos 65° + cos 25° sin 65°
Ans :
sin 25° cos( 90 - 25) + cos 25° sin ( 90 - 25)
⇒ sin 25° . sin 25° + cos 25° . cos 25°
2 2
⇒ sin 25° + cos 25°
⇒ 1 Ans . . . .
4. Choose the correct option . Justify your choice.
2 2
(i) 9 sec A - 9 tan A =
(A) 1 (B) 9
(C) 8 (D) 0
Ans : (B)
(ii) ( 1 + tanθ + secθ ) ( 1 + cotθ - cosecθ ) =
(A) 0 (B) 1
(C) 2 (D) -1
Ans : (C)
(iii) (secA + tanA ) ( 1 - sinA ) =
(A) sec A (B) sin A
(C) cosec A (D) cos A
Ans : (D)
2
1 + tan A
(iv) –––––––––––– =
2
1 + cot A
2
(A) sec A (B) -1
2 2
(C) cot A (D) tan A
Ans : (D)
5. Prove the following identities , where the angles involved are acute angles for which for which the expressions are defined.
2 1 - cos θ
(i) ( cosec θ - cot θ ) = ––––––––––
1 + cos θ
Ans :
2
L.H.S = ( cosecθ - cotθ )
2 2
= cosec θ + cot θ - 2 cosec θ cot θ
2
1 cos θ 1 cos θ
= ——— + ———— - 2 ——— . ———
2 2 sin θ sin θ
sin θ sin θ
2
1 cos θ 2 cos θ
= ——— + ———— - —————
2 2 2
sin θ sin θ sin θ
2
1 + cos θ - 2 cos θ
= ————————————
2
sin θ
2
( 1 - cos θ)
= ————————————
2
1 - cos θ
2
( 1 - cos θ)
= ———————————
( 1 - cos θ ) ( 1 + cos θ)
1 - cos θ
= ————— = R.H.S proved . . .
1 + cos θ
cos A 1 + sin A
(ii) ––––––––– + ––––––––––– = 2 sec A
1 + sin A cos A
Ans :
cos A 1 + sin A
L.H.S = ––––––––– + –––––––––––
1 + sin A cos A
2 2
cos A + ( 1 + sin A )
= ––––––––––––––––––––––
( 1 + sin A ) cos A
2 2
cos A + 1 + sin A + 2 sinA
= ––––––––––––––––––––––––––
( 1 + sin A ) cos A
1 + 1 + 2 sinA
= –––––––––––––––––––
( 1 + sin A ) cos A
2+ 2 sinA
= –––––––––––––––––––
( 1 + sin A ) cos A
2 ( 1 + sinA )
= –––––––––––––––––––
( 1 + sin A ) cos A
2
= –––––––––
cos A
= 2 sec A = R.H.S proved . . .
tan θ cot θ
(iii) –––––––––– + –––––––––– = 1 + sec θ cosec θ
1 - cot θ 1 - tan θ
Ans :
tan θ cot θ
L.H.S = ––––––––– + –––––––––
1 - cot θ 1 - tan θ
1
tan θ –––––––
tan θ
= ––––––––––– + ––––––––––––––
1 1 - tan θ
1 - –––––
tan θ
tan θ 1
= ––––––––––– + ––––––––––––––
tan θ - 1 tan θ ( 1 - tan θ )
––––––––––
tan θ
2
tan θ 1
= ––––––––––– + ––––––––––––––
tan θ - 1 tan θ ( 1 - tan θ )
2
tan θ 1
= ––––––––––– + –––––––––––––––––
tan θ - 1 tan θ { - ( tan θ - 1 )}
2
tan θ 1
= ––––––––––– - –––––––––––––––––
tan θ - 1 tan θ ( tan θ - 1 )
3
tan θ - 1
= ––––––––––––––––––
tan θ ( tan θ - 1 )
2
( tan θ - 1 ) ( tan θ + 1 + tan θ )
= ––––––––––––––––––––––––––––
tan θ ( tan θ - 1 )
sec θ + tan θ
= –––––––––––––––––
tan θ
2
sec θ tan θ
= ––––––– + –––––––––
tan θ tan θ
1
–––––– 2
cos θ
= –––––––––––– + 1
sin θ
––––––
cos θcos θ
= –––––––––––––– + 1
2
cos θ x sin θ
1
= ––––––––––––– + 1
sin θ cos θ
= 1 + cosec θ sec θ = R.H.S proved . . .
2
1 + secA sin A
(iv) ––––––––– = ––––––––––
secA 1 - cosA
Ans :
1 + secA
L.H.S = –––––––––
secA
1
1 + –––––––
cos A
= ––––––––––––––––––
1
–––––––––
cos A
cos A + 1
–––––––––––
cos A
= ––––––––––––––––––
1
–––––––––
cos A
( cos A + 1 ) cos A
= ––––––––––––––––––
cos A
= ( cos A + 1 )
1 - cos A
= ( 1 + cos A ) X –––––––––––––––
1 - cos A
2
1 - cos A
= –––––––––––––
1 - cos A
2
sin A
= ––––––––– = R.H.S proved . . .
1 - cos A
cosA - sinA + 1
(v) –––––––––––––––– = cosecA + cotA
cosA + sinA - 1
Ans :
cosA - sinA + 1
L.H.S = ––––––––––––––––
cosA + sinA - 1
cosA - ( sinA - 1 )
= ––––––––––––––––
cosA + ( sinA - 1 )
On rationalizing the denominator
cosA - ( sinA - 1 ) cosA - ( sinA - 1 )
= –––––––––––––––– X ––––––––––––––––
cosA + ( sinA - 1 ) cosA - ( sinA - 1 )
2
{ cosA - ( sinA - 1 )}
= –––––––––––––––––––––––
2 2
{ cosA } - { ( sinA - 1 ) }
2 2
cos A + ( sin A - 1 ) - 2 cos A ( sin A - 1 )
= –––––––––––––––––––––––––––––––––––––
2 2
{ cosA } - { ( sinA - 1 ) }
2 2
cos A + sin A + 1 - 2 sin A - 2 cos A sin A + 2 cos A
= ––––––––––––––––––––––––––––––––––––––––––––––
2 2
cos A - { sin A + 1 - 2 sin A }
1 + 1 - 2 sin A - 2 cos A sin A + 2 cos A
= –––––––––––––––––––––––––––––––––––––
2 2
cos A - sin A - 1 + 2 sin A
2 - 2 sin A - 2 cos A sin A + 2 cos A
= –––––––––––––––––––––––––––––––––––
2 2 2 2
cos A - sin A - ( cos A + sin A ) + 2 sin A
2 + 2 cos A - 2 sin A - 2 cos A sin A
= –––––––––––––––––––––––––––––––––––
2 2 2 2
cos A - sin A - cos A - sin A + 2 sin A
2 ( 1 + cos A ) - 2 sin A ( 1 + cos A )
= –––––––––––––––––––––––––––––––––––
2
- 2 sin A + 2 sin A
2 ( 1 + cos A ) ( 1 - sin A )
= –––––––––––––––––––––––––––
2 sin A (1 - sin A )
1 + cos A
= ––––––––––––
sin A
1 cos A
= –––––– + ––––––
sin A sin A
= cosec A + cot A = R.H.S proved . . .
3
sinθ - 2 sin θ
(vii) –––––––––––––– = tanθ
3
2cos θ - cosθ
Ans :
3
sinθ - 2 sin θ
L.H.S. = ––––––––––––––
3
2cos θ - cosθ
2
sinθ ( 1 - 2 sin θ )
= ––––––––––––––
2
cos θ( 2 cos θ - 1 )
2 2 2
sinθ ( sin θ + cos θ - 2 sin θ )
= –––––––––––––––––––––––––––
2 2 2
cos θ( 2 cos θ - sin θ + cos θ )
2 2
sinθ ( cos θ - sin θ )
= ––––––––––––––––––––
2 2
cos θ( cos θ - sin θ )
sinθ
= –––––– = tanθ = R.H.S proved . . .
cos θ
2 2 2 2
(viii) (sinA + cosecA) + (cosA + secA) = 7 + tan A + cot A
Ans :
2 2
L.H.S.= (sinA + cosecA) + (cosA + secA)
2 2 2 2
= sin A + cosec A + 2 sinA cosecA + cos A + sec A + 2 cosA secA
2 2 2 2
= sin A + cos A + sec A + 2 cosA secA + cosec A + 2 sinA cosecA
2 2
= 1 + sec A + 2 + cosec A + 2
2 2
= 5 + sec A + cosec A
2 2
= 5 + 1 + tan A + 1 + cot A
2 2
= 7 + tan A + cot A = R.H.S proved . . .
1
(ix) ( cosecA -sinA ) ( secA - cosA) = –––––––––––
tanA + cotA
Ans :
L.H.S = ( cosecA -sinA ) ( secA - cosA)
= cosecA secA - cosecA cosA - sinA secA + sinA cosA
1 1 1 1
= ––––– . ––––– - ––––– cosA - sin A ––––– + sinAcosA
sinA cosA sinA cosA
1 cosA sin A
= ––––––––– - ––––– - ––––– + sinA cosA
sinA cosA sinA cosA
2 2 2 2
1 - cos A - sin A + sin A cos A
= ––––––––––––––––––––––––––––––
sinA cosA
2 2 2 2
1 - ( cos A + sin A ) + sin A cos A
= ––––––––––––––––––––––––––––––
sinA cosA
2 2
1 - 1 + sin A cos A
= –––––––––––––––––––––––––
sinA cosA
2 2
sin A cos A
= ––––––––––––––––
1 - 2 sin A cosA
(vii) –––––––––––––– = tanθ
3
2cos θ - cosθ
Ans :
3
sinθ - 2 sin θ
L.H.S. = ––––––––––––––
3
2cos θ - cosθ
2
sinθ ( 1 - 2 sin θ )
= ––––––––––––––
2
cos θ( 2 cos θ - 1 )
sinθ ( sin θ + cos θ - 2 sin θ )
= –––––––––––––––––––––––––––
2 2 2
cos θ( 2 cos θ - sin θ + cos θ )
2 2
sinθ ( cos θ - sin θ )
= ––––––––––––––––––––
2 2
cos θ( cos θ - sin θ )
sinθ
= –––––– = tanθ = R.H.S proved . . .
cos θ
(viii) (sinA + cosecA) + (cosA + secA) = 7 + tan A + cot A
Ans :
2 2
L.H.S.= (sinA + cosecA) + (cosA + secA)
2 2 2 2
= sin A + cosec A + 2 sinA cosecA + cos A + sec A + 2 cosA secA
2 2 2 2
= sin A + cos A + sec A + 2 cosA secA + cosec A + 2 sinA cosecA
2 2
= 1 + sec A + 2 + cosec A + 2
2 2
= 5 + sec A + cosec A
2 2
= 5 + 1 + tan A + 1 + cot A
2 2
= 7 + tan A + cot A = R.H.S proved . . .
1
(ix) ( cosecA -sinA ) ( secA - cosA) = –––––––––––
tanA + cotA
Ans :
L.H.S = ( cosecA -sinA ) ( secA - cosA)
= cosecA secA - cosecA cosA - sinA secA + sinA cosA
1 1 1 1
= ––––– . ––––– - ––––– cosA - sin A ––––– + sinAcosA
sinA cosA sinA cosA
1 cosA sin A
= ––––––––– - ––––– - ––––– + sinA cosA
sinA cosA sinA cosA
2 2 2 2
1 - cos A - sin A + sin A cos A
= ––––––––––––––––––––––––––––––
sinA cosA
1 - ( cos A + sin A ) + sin A cos A
= ––––––––––––––––––––––––––––––
sinA cosA
1 - 1 + sin A cos A
= –––––––––––––––––––––––––
sinA cosA
sin A cos A
= ––––––––––––––––
sinA cosA
= sinA cosA
1
R.H.S. = –––––––––––
tanA + cotA
1
= –––––––––––––––––––––
sinA cosA
–––––––– + –––––––
cosA sinA
1
= –––––––––––––––––––––
2 2
sin A+cos A
–––––––––––––––
cosA sinA
= cosA sinA
therfore L.H.S. = R.H.S. proved . . .
2 2
( 1 + tan A ) ( 1 - tan A ) 2
(x) ––––––––––––– = –––––––––––––– = tan A
2 2
( 1 + cot A ) ( 1 - cot A )
Ans :
2
( 1 + tan A )
–––––––––––––
2
( 1 + cot A )
2
sec A
= ––––––––––
2
cosec A
2
= tan A
2
( 1 - tan A )
–––––––––––––
2
( 1 - cot A )
2
1 + tan A - 2 tan A
= –––––––––––––––––––––
2
1 + cot A - 2 cot A
2
sec A - 2 tan A
= –––––––––––––––––
2
cosec A - 2 cot A
1 2 sin A
–––––– - ––––––––––
2 cos A
cos A
= –––––––––––––––––––––
1 2 cos A
––––––– - –––––––––
2 sin A
sin A
1 - 2 sin A cosA
–––––––––––––––––––––
2
cos A
= ––––––––––––––––––––––––––––
1 - 2 sin A cos A
–––––––––––––––––––––
2
sin A
2
sin A ( 1 - 2 sin A cos A )
= –––––––––––––––––––––––––
2
cos A ( 1 - 2 sin A cos A )
2
= tan A
Hence
2 2
( 1 + tan A ) ( 1 - tan A ) 2
––––––––––––––––– = –––––––––––––––= tan A
2 2
( 1 + cot A ) ( 1 - cot A )
proved . . . .
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