Question

root 1+sin tita /1-sin teta + root 1-sin teta 1+ sin teta​

Answer 1

Answer:-

To Prove:

$\sf \implies \large{ \sqrt{ \frac{1 + \sin \theta}{1 - \sin \theta} } + \sqrt{ \frac{ 1 - \sin \theta}{1 + \sin \theta} } = 2 \: \sec \theta}$

On squaring both sides we get,

$\sf \implies \: \bigg( { \sqrt{ \frac{1 + \sin \theta }{1 - \sin \theta} } + \sqrt{ \frac{1 - \sin \theta }{1 + \sin \theta } } \bigg)}^{2} = {(2 \: \sec \theta) }^{2}$

Using the formula (a + b)² = a² + b² + 2ab in LHS we get,

$\sf \implies \: { \bigg( \sqrt{ \frac{1 + \sin \theta }{1 - \sin \theta} } \bigg)}^{2} + { \bigg( \sqrt{ \frac{1 - \sin \theta }{1 + \sin \theta} } \bigg) } ^{2} + 2 \times \sqrt{ \frac{1 + \sin \theta}{1 - \sin \theta } \times \frac{1 - \sin \theta }{1 + \sin \theta } } = 4 \: { \sec}^{2} \theta \\ \\ \sf \implies \: \frac{1 + \sin \theta }{1 - \sin \theta } + \frac{1 - \sin \theta }{1 + \sin \theta } + 2 = 4 \: { \sec}^{2} \theta \\ \\ \sf \implies \: \frac{{(1 + \sin \theta)} ^{2} + ( {1 - \sin \theta)}^{2} }{(1 + \sin \theta )(1 - \sin \theta } = 4 \: { \sec }^{2} \theta - 2$

Using the formulae

• (a + b)(a - b) = a² - b²

• (a - b)² = a² + b² - 2ab

in LHS we get,

$\sf \implies \: \frac{1 + { \sin }^{2} \theta + 2 \sin \theta + 1 + { \sin}^{2} \theta - 2 \sin \theta}{ {1}^{2} - { \sin}^{2} \theta} = 2(2 \: { \sec }^{2} \theta \: - 1) \\ \\ \sf \implies \: \frac{2 (1 + { \sin }^{2} \theta) }{1 - { \sin }^{2} \theta } =2(2 \: { \sec } ^{2} \theta - 1)$

Using the identity cos² A = 1 - sin² A in LHS we get,

$\sf \frac{1 }{ { \cos}^{2} \theta } + \frac{ { \sin }^{2} \theta }{ {\cos} ^{2} \theta } = 2 \: { \sec }^{2} \theta - 1$

Using the identies sin² A/cos² A = tan² A and 1/cos² A = sec² A in LHS we get,

$\sf \implies \: { \sec}^{2} \theta + { \tan }^{2} \theta = 2 \: { \sec }^{2} \theta - 1$

We know that,

sec² A - tan² A = 1

→ sec² A - 1 = tan² A

Hence,

$\sf \implies \: { \sec }^{2} \theta + { \sec }^{2} \theta - 1 = 2 { \sec }^{2} \theta - 1 \\ \\ \sf \implies \: 2 { \sec }^{2} \theta - 1 = 2 { \sec }^{2} \theta - 1 \\ \\ \sf \large{LHS \: = RHS}$

Hence, Proved.