Question

cos theta by 1- tan theta + sin theta by 1-cot theta =cos theta + sin theta ​

Answer 1

Question :-

Prove that

$\frac{ \cos \theta}{1 - \tan \theta} + \frac{ \sin \theta}{1 - \cot \theta} = \cos \theta \: + \sin \theta \:$

Proof :-

Formula used :-

$\boxed{ \star }\: \: \: \tan \theta \: = \frac{ \sin \theta}{ \cos \theta} \\ \\ \boxed{\star } \: \: \cot \theta \: = \frac{ \cos \theta}{ \sin \theta} \\ \\ \boxed{ \star} \: \: \: \sf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}$

Explanation :-

Firstly take left hand side ( LHS)

$\rightarrow \: \frac{ \cos \theta}{1 - \tan \theta} + \frac{ \sin \theta}{1 - \cot \theta} \: \\ \\ \rightarrow \: \frac{ \cos \theta}{1 - \frac{ \sin \theta}{ \cos \theta} } + \frac{ \sin \theta}{1 - \frac{ \cos \theta}{ \sin \theta \: } } \\ \: \\ \rightarrow \: \frac{ \cos \theta}{ \frac{ \cos \theta \: - \sin \theta}{ \cos \theta} } + \frac{ \sin \theta}{ \frac{ \sin \theta \: - \cos \theta \: }{ \sin \theta} } \\ \\ \rightarrow \: \frac{ { \cos}^{2} \theta}{ \cos \theta \: - \sin \theta \: } + \frac{ { \sin}^{2} \theta}{ \sin \theta \: - \cos \theta} \\ \\ \sf{ taking \:common \: negetive \: sign \: } \\ \\ \rightarrow \: \frac{ { \cos}^{2} \theta}{ \cos \theta - \sin \theta} - \frac{ { \sin}^{2} \theta}{ \cos \theta \: - \sin \theta \: } \\ \\ \rightarrow \: \frac{ { \cos}^{2} \theta - { \sin}^{2} \theta }{ \cos \theta - \sin \: \theta \: } \\ \\ \because \sf{{x}^{2} - {y}^{2} = (x + y)(x - y)} \\ \therefore \: \\ \\ \rightarrow \: \frac{( \cos \theta - \sin \theta)( \cos \theta \: + \sin \theta)}{ \cos \theta - \sin \theta} \\ \\ \rightarrow \: \cos \theta \: + \sin \theta \:$

LHS = RHS

left hand side = right hand side

hence proved : )

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