Question

show that cos 0/1-sin0 +cos0/1-sin0 = 2sec0​

Answer 1

CORRECT QUESTION :-

$\\ \red \bigstar \displaystyle \tt \: prove \: that \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta$

GIVEN :-

$\\ \red \bigstar \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta$

TO PROVE :-

$\\ \red \bigstar \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta$

SOLUTION :-

$\\ \red \bigstar \: \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta$

Take L.H.S Part,

$\\ \implies \displaystyle \tt \: \dfrac{ \cos \theta }{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta }$

$\blue \bigstar \orange{ \tt \: Taking \: L.C.M \bigg(1 + \sin \theta\bigg)\bigg(1 - \sin \theta\bigg)}$

$\implies \displaystyle \tt \: \dfrac{ \cos \theta\bigg(1 - \sin \theta \bigg) + \cos \theta \bigg(1 + \sin \theta \bigg)} { \bigg(1 + \sin \theta \bigg) \bigg(1 - \sin \theta \bigg) }$

$\displaystyle \tt \implies \: \dfrac{ \cos \theta - \cos \theta. \sin \theta + \cos \theta + \cos \theta. \sin \theta}{\bigg(1 + \sin \theta\bigg)\bigg(1 - \sin \theta\bigg)}$

$\blue \bigstar \orange{\tt Using \: identity :- (a + b)(a -b) = a^{2} - b^{2}}$

$\implies \displaystyle \tt \: \dfrac{ \cos \theta + \cos \theta}{(1) ^{2} - \sin ^{2} \theta }$

$\implies \displaystyle \tt \: \dfrac{2 \cos \theta}{1 - \sin ^{2} \theta }$

$\blue \bigstar \orange{\tt Using \: identity :- 1 - \sin ^{2} \theta = \cos ^{2} \theta }$

$\implies \displaystyle \tt \: \dfrac{2\cos \theta}{ \cos ^{2}\theta}$

$\implies \displaystyle \boxed{\tt \:2 \sec \theta}$

L.H.S = R.H.S

HENCE VERIFIED ✅