Question

Prove that LHS=RHS......,......

Answer 1

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$\sqrt{ { \sec( \alpha ) }^{2} + { \csc( \alpha ) }^{2} } \\ \sqrt{(1 + { \tan( \alpha ) }^{2} ) + (1 + { \cot( \alpha ) }^{2}) } \\ \sqrt{2 + { \tan( \alpha ) }^{2} + { \cot( \alpha ) }^{2} } \\ \sqrt{2 \tan( \alpha ) \cot( \alpha ) + { \tan( \alpha ) }^{2} + { \cot( \alpha ) }^{2} } \\ \sqrt{ {( \tan( \alpha ) + \cot( \alpha ) ) }^{2} } \\ = \tan( \alpha ) + \cot( \alpha )$
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Answer 2

Given:

$\sqrt{ {\sec}^{2}\theta + {\cosec}^{2}\theta } = \tan\theta + \cot\theta$

To Prove:

LHS = RHS

Answer:

We will be converting every term in sin $\theta$ and cos $\theta$.

$\bfWe \:will \:solve\: LHS\: to\: be\: equal\: to\: RHS.$

LHS:

We know that,

$\sec\theta = \frac{1}{\cos\theta} \: \: \: \: \\ \\ and \\ \\ \csc\theta = \frac{1}{\sin\theta}$

Substituting the value of sec $\theta$ and cosec $\theta$ in the equation:

$\sqrt{\frac{1}{\cos^{2}\theta } + \frac{1}{ {\sin}^{2} \theta} }$

Taking LCM, we get:

$\sqrt{ \frac{ {\sin}^{2}\theta + {\cos}^{2}\theta }{ {\cos}^{2}\theta \: {\sin}^{2}\theta } }$

Substituting the value of ${\sin}^2\theta+{\cos}^2\theta$ as 1.

$\sqrt{ \frac{1}{ {\cos}^{2}\theta \: {\sin}^{2} \theta} }$

Taking square root, we get:

$\frac{1}{\cos\theta \: \sin\theta}$

Substituting the value of 1 as ${\sin}^2\theta+{\cos}^2\theta$ in the above equation.

$\frac{ {\sin}^{2}\theta \: + \: {\cos}^{2}\theta}{\cos\theta \: \sin\theta}$

On dividing , we get:

$\frac{ {\sin}^{2}\theta }{\cos\theta\sin\theta} + \frac{ {\cos}^{2}\theta}{\cos\theta\sin\theta } \\ \\ \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tan\theta + \cot\theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence Proved

⭐Other Trigonometric Identities:⭐

$\\\\1) \: {\csc}^{2} \theta - {\cot}^{2} \theta = 1$

$2) \: {\sec}^{2} \theta - {\tan}^{2} \theta = 1$

⭐Trigonometric Ratios:⭐

$\\\\1) \: \sin\theta = \frac{1}{ \cosec \theta} \\ 2) \: \cos \: \theta \: \: = \: \frac{1}{\sec\theta} \\ 3) \: \tan \: \theta \: = \: \frac{1}{\cot \: \theta} \\ 4) \: \cot \: \theta \: = \: \frac{1}{\tan \: \theta} \\ 5) \: \sec \: \theta \: = \: \frac{1}{\cos \: \theta} \\ 6) \:\csc\:\theta= \frac{1}{\sin \: \theta}$