Question

Use the
EXAMPLE 1. Prove the following
1+ sin e
(i)
sec 0 + tan o (ii)
Vi-sin e
et​

Answer 1

AnswEr :

Given Expression,

$\sf \: \sqrt{ \dfrac{1 + sin \: x}{1 - sin \: x} } = sec \: x + tan \: x$

LHS

$\sf \: \sqrt{ \dfrac{1 + sin \: x}{1 - sin \: x} }$

Rationalizing the denominator,we get :

$\implies \sf \: \sqrt{ \dfrac{1 + sin \: x}{1 - sin \: x} \times \dfrac{1 + sin \: x}{1 + sin \: x} } \\ \\ \implies \: \sf \: \sqrt{ \dfrac{ \: \: ( 1 + sin \: x) {}^{2} }{1 - sin {}^{2}x } }$

We know that,

sin²x + cos²x = 1 ➠ 1 - sin²x = cos²x

$\implies \: \sf \: \sqrt{ \dfrac{(1 + sin \: x) {}^{2} }{cos {}^{2} x} } \\ \\ \implies \: \sf \: \sqrt{ \bigg( \dfrac{1 + sin \: x}{cos \: x} \bigg) {}^{2} } \\ \\ \implies \: \sf \: \dfrac{1}{cos \: x} + \dfrac{sin \: x}{cos \: x} \\ \\ \implies \: \sf \: sec \: x + tan \: x \implies \: RHS$

Answer 2

Question :

Prove : $\tt {\sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} } = sec \: \theta + tan \: \theta}$

Answer :

Solving LHS :

$\sf { \sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} } }$

Rationalizing :

$\leadsto \sf { = \sqrt{ \dfrac{1 + sin \: \theta}{1 - sin \: \theta} \times \dfrac{1 + sin \: \theta}{1 + sin \: \theta} }}$

$\leadsto \sf {= \sqrt{ \dfrac{ {( 1 + sin \: \theta)}^{2} }{1 - {sin}^{2}\theta } }}$

Identity : sin²∅ + cos²∅ = 1

$\leadsto { \sf {=\sqrt{ \dfrac{{(1 + sin \: \theta)}^{2} }{{cos}^{2} \theta} }} }$

$\leadsto \sf {= \sqrt{{ \bigg( \dfrac{1 + sin \: \theta}{cos \: \theta} \bigg)}^{2} }}$

$\leadsto \sf{ =\dfrac{1}{cos \: \theta} + \dfrac{sin \: \theta}{cos \: \theta} }$

$\leadsto \sf {=sec \: \theta + tan \: \theta }$

Hence, Proved.