# Prove the following identities, where the angles involved are acute angles for which the expressions are defined :

## Question :

(i)  $$1 – cos\theta\over 1 + cos\theta$$ = $$({cosec\theta – cot\theta})^2$$

(ii)  $$1 + sin\theta\over cos\theta$$ + $$cos\theta\over 1 + sin\theta$$ = $$2sec\theta$$

(iii)  $$tan\theta\over 1 – cot\theta$$ + $$cot\theta\over 1 – tan\theta$$ = $$sec\theta + cosec\theta$$ + 1

(iv)  $$1 + sec\theta\over sec\theta$$ = $$sin^2\theta\over 1 – cos\theta$$

(v)  $$cos A – sin A + 1\over cos A + sin A – 1$$ = cosec A + cot A

(vi)  $$\sqrt{1 + sin\alpha\over 1 – sin\alpha}$$ = $$sec\alpha + tan\alpha$$

(vii)  $$sin\theta – 2 sin^3\theta\over 2cos^3\theta – cos\theta$$ = $$tan\theta$$

(viii)  $$({sin A + cosec A})^2$$ + $$({cos A + sec A})^2$$ = 7 + $$tan^2 A$$ + $$cot^2A$$

(ix)  (cosec A – sin A)(sec A – cos A) = $$1\over tan A + cot A$$

(x)  $$1 + tan^2\theta\over 1 + cot^2\theta$$ = $$({1 – tan\theta\over 1 – cot\theta})^2$$

## Solution :

(i)  R.H.S = $$({cosec\theta – cot\theta})^2$$ = $$({1\over sin\theta} – {cos\theta\over sin\theta})^2$$

= $$({1 – cos\theta\over sin\theta})^2$$ = $$({1 – cos\theta})^2\over sin^2\theta$$ = $$({1 – cos\theta})^2\over 1 – cos^2\theta$$

= $$1 – cos\theta\over 1 + cos\theta$$ = L.H.S

(ii)  L.H.S = $$1 + sin\theta\over cos\theta$$ + $$cos\theta\over 1 + sin\theta$$ = $$({1 + sin\theta})^2 + cos^2\theta\over cos\theta(1 + sin\theta)$$

= $$1 + sin^2\theta + 2sin\theta + cos^2\theta\over cos\theta(1 + sin\theta)$$

= $$1 + 1 + 2sin\theta\over cos\theta(1 + sin\theta)$$ = $$2(1 + sin\theta)\over cos\theta(1 + sin\theta)$$ = $$2\over cos\theta$$

= $$2\sec\theta$$ = R.H.S.

(iii)  L.H.S = $$tan\theta\over 1 – cot\theta$$ + $$cot\theta\over 1 – tan\theta$$ = $$tan\theta\over 1 – {1\over tan\theta}$$ + $${1\over tan\theta}\over 1 – tan\theta$$

= $$tan^2\theta\over tan\theta – 1$$ – $$1\over tan\theta(tan\theta – 1)$$ = $$tan^3\theta – 1\over tan\theta(tan\theta – 1)$$

= $${(tan\theta – 1)(tan^2\theta + tan\theta + 1)}\over tan\theta(tan\theta – 1)$$ = $$sec^2\theta + tan^2\theta\over tan\theta$$

= $$sec^2\theta\over tan\theta$$ + $$tan\theta\over tan\theta$$ = $$sec\theta cosec\theta$$ + 1 = R.H.S

(iv)  L.H.S = $$1 + sec\theta\over sec\theta$$ = $$cos\theta + 1\over cos\theta$$ $$\times$$ $$cos\theta\over 1$$

= $$cos\theta + 1$$ = $$(1 + cos\theta)(1 – cos\theta)\over 1 – cos\theta$$

= $$1 – cos^2\theta\over 1 – cos\theta$$ = $$sin^2\theta\over 1 – cos\theta$$ = R.H.S.

(v)  L.H.S = $$cos A – sin A + 1\over cos A + sin A – 1$$

Dividing numerator and denominator by sin A, we get

L.H.S = $$cot A + cosec A – 1\over 1 – cosec A + cot A$$ = $$cot A + cosec A – (cosec^2 A – cot^2A)\over 1 – cosec A + cot A$$

= $$cot A + cosec A – (cosec A + cot A)(cosec A – cot A)\over 1 – cosec A + cot A$$

= $$(cot A + cosec A)(1 – cosec A + cot A)\over 1 – cosec A + cot A$$ = cot A + cosec A = R.H.S

(vi)  L.H.S = $$\sqrt{1 + sin\alpha\over 1 – sin\alpha}$$ = $$\sqrt{(1 + sin\alpha)(1 + sin\alpha)\over (1 – sin\alpha)(1 + sin\alpha)}$$

= $$\sqrt{(1 + sin\alpha)^2\over 1 – sin^2\alpha}$$ = $$\sqrt{(1 + sin\alpha)^2\over cos^2\alpha}$$

= $$1 + sin\alpha\over cos\alpha$$ = $$1\over cos\alpha$$ + $$sin\alpha\over cos\alpha$$ = $$sec\alpha + tan\alpha$$ = R.H.S

(vii)  L.H.S = $$sin\theta – 2 sin^3\theta\over 2cos^3\theta – cos\theta$$ = $$sin\theta(1 – 2sin^2\theta)\over cos\theta(2cos^2\theta – 1)$$

= $$sin\theta[1 – 2(1 – cos^2\theta)]\over cos\theta[2cos^2\theta – 1]$$ = $$sin\theta[1 – 2 + 2cos^2\theta]\over cos\theta[2cos^2\theta – 1]$$

= $$sin\theta[2cos^2\theta – 1]\over cos\theta[2cos^2\theta – 1]$$ = $$sin\theta\over cos\theta$$ = $$tan\theta$$ = R.H.S.

(viii)  L.H.S = $$({sin A + cosec A})^2$$ + $$({cos A + sec A})^2$$

= $$sin^2A$$ + $$cosec^2A$$ + 2sin A cosec A + $$cos^2A$$ + $$sec^2A$$ + 2cos A.sec A

= $$sin^2A$$ + $$cosec^2A$$ + $$2sin A\over sin A$$  + $$cos^2A$$ + $$sec^2A$$ + $$2cos A\over cos A$$

= $$sin^2A$$ + $$cosec^2A$$  + $$cos^2A$$ + $$sec^2A$$ + 4

= $$sin^2A$$ + $$cos^2A$$ + $$cosec^2A$$ + $$sec^2A$$ + 4

= 1 + (1 +$$cot^2A$$) + (1 + $$tan^2A$$) + 4

= 7 + $$cot^2A$$ + $$tan^2A$$ = R.H.S

(ix)  L.H.S = (cosec A – sin A)(sec A – cos A)

= cosec A sec A – cosec A cos A – sin A sec A + sin A cos A

= $$1\over sin A cos A$$ – $$cos A\over sin A$$ – $$sin A\over cos A$$ + sin A cos A

= $$1 – cos^2A – sin^2A + sin^2A.cos^2A\over sin A.cos A$$

= $$1 – (cos^2A + sin^2A) + sin^2A.cos^2A\over sin A.cos A$$ = = $$1 – 1 + sin^2A.cos^2A\over sin A.cos A$$

= = $$sin^2A.cos^2A\over sin A.cos A$$ = sin A cos A                       ………..(1)

R.H.S = $$1\over tan A + cot A$$ = $$1\over {sin A\over cos A} + {cos A\over sin A}$$

= $$sin A cos A\over sin^2A + cos^2A$$ = sin A cos A                       …………(2)

From (1) and (2), we have

L.H.S = R.H.S

(x)  L.H.S = $$1 + tan^2\theta\over 1 + cot^2\theta$$ = $$sec^2\theta\over cosec^2\theta$$

= $$sin^2\theta\over cos^2\theta$$ = $$tan^2\theta$$      ……….(1)

R.H.S = $$({1 – tan\theta\over 1 – cot\theta})^2$$ = $$({1 – tan\theta\over 1 – {1\over tan\theta}})^2$$

= $$({1 – tan\theta\over {tan\theta – 1\over tan\theta}})^2$$ = $$({1 – tan\theta\over -(1 – tan\theta)}.tan\theta)^2$$ = $$tan^2\theta$$               ……(2)

From (1) and (2), clearly, L.H.S = R.H.S