Question

Prove that question ​

Answer 1

Question :-

prove that :

$\frac{ \tan \theta}{1 + \sec \theta} + \frac{1 + \sec \theta}{ \tan \theta \: } = 2 \csc \theta$

Formula used :-

$\star \: \: { \sec}^{2} \theta - { \tan}^{2} \theta = 1 \\ \\ \star \: \frac{1}{ \cos \theta} = \sec \theta \\ \\ \star \: \: \frac{1}{ \sin \theta} = \csc \theta \\ \\ \star \: \: \frac{ \sin \theta}{ \cos \theta} = \tan \theta$

Explanation :-

Taking left hand side (LHS)

$\rightarrow \: \frac{ \tan \theta}{1 + \sec \theta} + \frac{1 + \sec \theta}{ \tan \theta \: } \: \\ \\ \rightarrow \frac{ { \tan}^{2} \theta + {(1 + \sec \theta)}^{2} }{ \tan \theta(1 + \sec \theta)} \\ \\ \rightarrow \: \frac{ { \tan}^{2} \theta + 1 + { \sec}^{2} \theta + 2 \sec \theta }{tan \theta(1 + \sec \theta)} \\ \\ \rightarrow \: \frac{ { \sec}^{2} \theta - 1 + 1 + { \sec}^{2} \theta + 2 \sec \theta }{ \tan \theta(1 + \sec \theta)} \\ \\ \rightarrow \: \frac{2 { \sec}^{2} \theta + 2 \sec \theta }{ \tan \theta(1 + \sec \theta)} \\ \\ \rightarrow \: \frac{2 \sec \theta( \sec \theta + 1)}{ \tan \theta(1 + \sec \theta)} \\ \\ \rightarrow \: \frac{2 \sec \theta}{ \tan \theta} \\ \\ \rightarrow \: \frac{ \frac{2}{ \cos \theta} }{ \frac{ \sin \theta}{ \cos \theta } } \\ \\ \rightarrow \: \: \frac{2}{ \sin \theta} \\ \\ \rightarrow \: 2 \csc \theta \:$

LHS = RHS

hence proved .

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