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

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