Question

Trigeonometric Identities :

Exercise 5.1

Question 9
Cos x/1-sin x = 1 + cos x + sin x/ 1+ cos x- sinx
If you dont know the answer so stay away !!

Answer 1

Solution :

Let's considered RHS :

$\: \: \: \: \: \rightarrow \tt \large{\frac{1 + \cos \: x + \sin \: x }{1 + \cos \: x - \sin \: x } }$

$\: \: \: \: \: \: \rightarrow \tt \large\frac{(1 + \cos \: x) + ( \sin \: x) }{(1 + \: \cos \: x) - ( \sin \: x) }$

$\tt \large \frac{(1 + \cos \: x) + ( \sin \: x) }{(1 + \cos \: x) - ( \sin \: x) } \times \frac{(1 + \cos \: x) + ( \sin \: x) }{(1 + \cos \: x) + ( \sin \: x) }$

$\: \: \: \: \: \: \rightarrow \tt \large \frac{[(1 + \cos \: x) + ( \sin \: x) ] {}^{2} }{(1 + \cos \: x) {}^{2} - ( \sin \: x) {}^{2} }$

$\: \: \: \: \: \: \: \: \large \rightarrow \tt \frac{[(1 + \cos \: x) + ( \sin \: x) ] {}^{2} }{(1 + \cos \: x) {}^{2} - ( \sin \: x) {}^{2} }$

$\tt \large \frac{(1 + \cos \: x) {}^{2} + ( \sin \: x) {}^{2} + 2(1 + \cos \: x) ( \sin \: x) }{(1 + \cos {}^{2}x + 2 \cos \: x ) - ( \sin {}^{2}x ) }$

$\sf \tt \frac{(1 + \cos \: x) {}^{2} + ( \sin \: x) {}^{2} + 2(1 + \cos \: x) ( \sin \: x) }{(1 + \cos \: {}^{2} x + 2 \cos \: x) - ( \sin {}^{2} x) }$

$\tt \large \frac{(1 + \cos \: x) {}^{2} + ( \sin \: x) {}^{2} + 2(1 + \cos \: x) ( \sin \: x) }{(1 + \cos {}^{2} x + 2 \cos \: x) - ( \sin {}^{2} x) }$

$\large \tt \frac{1 + \cos{}^{2} x + 2 \cos x + \sin {}^{2}x + 2 \sin x + 2 \sin x \: cos \: x }{1 + \cos {}^{2}x + 2 \cos x - sin {}^{2}x }$

$\sf \: we \: know \: { \sin }^{2} x + { \cos }^{2} x = 1$

$\large \tt \frac{1 + 1 + 2 \cos x + 2 \sin \: x + 2 \sin x \cos x }{(1 - {sin}^{2}x) + {cos}^{2}x + 2 \cos x }$

$\sf we \: know \: 1 - {cos}^{2} x = {sin}^{2}x$

$\large \tt \frac{2 + 2 \cos x + 2 \: sin \: x + 2 \:sin \: x \: cos \: x}{ {cos}^{2}x + {cos}^{2} x + 2 \cos x }$

$\large \tt\frac{2 + 2 \cos x + 2 \sin x + 2 \sin x \cos x }{2 \: {cos}^{2}x + 2 \cos \: x }$

$\tt \large \frac{2 + 2 \cos x + 2 \sin x + 2 \sin x \cos x }{ {cos}^{2}x + {cos}^{2} x + 2 \cos x }$

$\large \tt \frac{1 + \cos x + \sin x + \sin x \cos x }{ \cos x( \cos x + 1) }$

$\large \tt \frac{1(1 + \cos x) + \sin x( \cos x + 1) }{ \cos x( \cos x + 1) }$

$\large \tt \frac{(1 + \sin x)( \cos x + 1) }{ \cos x( \cos x + 1) }$

$\large \tt \frac{1 + \sin x }{cos \: x} \times \frac{cos \:x}{cos \: x}$

$\large \tt \frac{(1 + \sin x) \cos x }{ {cos}^{2} x}$

$\sf \: we \: know \: 1 - \sin {}^{2}x = {cos}^{2} x$

$\tt \large \frac{(1 + \sin x ) \cos x }{1 - {sin}^{2}x }$

$\tt \large \frac{(1 \: + \: sin \: x)cos \: x}{(1 - sin \: x)(1 + sin \: x)}$

$\large \tt \frac{cos \: x}{1 - sin \: x}$

$\tt= LHS$

$\tt∴ LHS = RHS$

$\tt \: Hence \: Proved$

Answer 2

Solution :

Let's considered RHS :

$\: \: \: \: \: \rightarrow \tt \large{\frac{1 + \cos \: x + \sin \: x }{1 + \cos \: x - \sin \: x } }→$

$</p><p>\: \: \: \: \: \: \rightarrow \tt \large\frac{(1 + \cos \: x) + ( \sin \: x) }{(1 + \: \cos \: x) - ( \sin \: x) }→$

$\tt \large \frac{(1 + \cos \: x) + ( \sin \: x) }{(1 + \cos \: x) - ( \sin \: x) } \times \frac{(1 + \cos \: x) + ( \sin \: x) }{(1 + \cos \: x) + ( \sin \: x) }$

$\: \: \: \: \: \: \rightarrow \tt \large \frac{[(1 + \cos \: x) + ( \sin \: x) ] {}^{2} }{(1 + \cos \: x) {}^{2} - ( \sin \: x) {}^{2} }→$

$\: \: \: \: \: \: \: \: \large \rightarrow \tt \frac{[(1 + \cos \: x) + ( \sin \: x) ] {}^{2} }{(1 + \cos \: x) {}^{2} - ( \sin \: x) {}^{2} }→$

$\tt \large \frac{(1 + \cos \: x) {}^{2} + ( \sin \: x) {}^{2} + 2(1 + \cos \: x) ( \sin \: x) }{(1 + \cos {}^{2}x + 2 \cos \: x ) - ( \sin {}^{2}x ) }$

$</p><p>\sf \tt \frac{(1 + \cos \: x) {}^{2} + ( \sin \: x) {}^{2} + 2(1 + \cos \: x) ( \sin \: x) }{(1 + \cos \: {}^{2} x + 2 \cos \: x) - ( \sin {}^{2} x) }$

$\tt \large \frac{(1 + \cos \: x) {}^{2} + ( \sin \: x) {}^{2} + 2(1 + \cos \: x) ( \sin \: x) }{(1 + \cos {}^{2} x + 2 \cos \: x) - ( \sin {}^{2} x) }$

$</p><p>\large \tt \frac{1 + \cos{}^{2} x + 2 \cos x + \sin {}^{2}x + 2 \sin x + 2 \sin x \: cos \: x }{1 + \cos {}^{2}x + 2 \cos x - sin {}^{2}x }$

$</p><p>\sf \: we \: know \: { \sin }^{2} x + { \cos }^{2}$

$\large \tt \frac{1 + 1 + 2 \cos x + 2 \sin \: x + 2 \sin x \cos x }{(1 - {sin}^{2}x) + {cos}^{2}x + 2 \cos x } </p><p>$

$\large \tt \frac{2 + 2 \cos x + 2 \: sin \: x + 2 \:sin \: x \: cos \: x}{ {cos}^{2}x + {cos}^{2} x + 2 \cos x } </p><p>.$

$\tt \large \frac{2 + 2 \cos x + 2 \sin x + 2 \sin x \cos x }{ {cos}^{2}x + {cos}^{2} x + 2 \cos x }$

$\large \tt \frac{1 + \cos x + \sin x + \sin x \cos x }{ \cos x( \cos x + 1) }$

$\large \tt \frac{1(1 + \cos x) + \sin x( \cos x + 1) }{ \cos x( \cos x + 1) }$

$\large \tt \frac{(1 + \sin x)( \cos x + 1) }{ \cos x( \cos x + 1) }$

$\large \tt \frac{1 + \sin x }{cos \: x} \times \frac{cos \:x}{cos \: x}$

$\large \tt \frac{(1 + \sin x) \cos x }{ {cos}^{2} x}$

$\sf \: we \: know \: 1 - \sin {}^{2}x = {cos}^{2}$

$\tt \large \frac{(1 + \sin x ) \cos x }{1 - {sin}^{2}x }$

$\tt \large \frac{(1 \: + \: sin \: x)cos \: x}{(1 - sin \: x)(1 + sin \: x)}$

$\large \tt \frac{cos \: x}{1 - sin \: x}$

=LHS

∴LHS=RHS

HenceProved