Question

Pls give the solve..​

Answer 1

❐ FULL EXPLANATION:

.

$\frac{ \sec(\theta) - \tan(\theta) }{ \sec(\theta) + \tan(\theta) } = \frac{36}{49} \: \: \: ...(given)$

So, We Can Say That,

$\sec( \theta) - \tan(\theta) = 36 \: \: \: ...(1)$

$\sec( \theta) + \tan(\theta) = 49 \: \: \: ...(2)$

So,

$\implies \frac{ \sec(\theta) - \tan(\theta) }{ \sec(\theta) + \tan(\theta) } = \frac{36}{49}$

$\implies \frac{ \sec(\theta) - \tan(\theta) + \sec(\theta) + \tan(\theta) }{ \sec(\theta) + \tan(\theta) } = \frac{36 + 49}{49} \: \: \: [by \: Eq \: (2)]$

$\implies \frac{ 2\sec(\theta) }{ \sec(\theta) + \tan(\theta) - (\sec(\theta) - \tan(\theta))} = \frac{85}{49 - 36} \: \: \: [by \: Eq \: (1)]$

$\implies \frac{ 2\sec(\theta) }{ \sec(\theta) + \tan(\theta) - \sec(\theta) + \tan(\theta)} = \frac{85}{13}$

$\implies \frac{ 2\sec(\theta) }{ 2 \tan(\theta)} = \frac{85}{13}$

$\implies \frac{ \sec(\theta) }{ \tan(\theta)} = \frac{85}{13}$

$\implies \frac{ \frac{1}{ \cos(\theta) } }{ \frac{ \sin(\theta) }{ \cos(\theta) } } = \frac{85}{13}$

$\bigg[ \because \: \sec( \theta) = \frac{1}{ \cos(\theta) } \: \: \: and \: \: \: \tan(\theta) = \frac{ \sin(\theta) }{ \cos(\theta) } \: \bigg]$

$\implies { \frac{1}{ \cos(\theta) } } \times { \frac{ \cos(\theta) }{ \sin(\theta) } } = \frac{85}{13}$

$\implies { \frac{1}{ \sin(\theta) } } = \frac{85}{13}$

$\implies \boxed {\cosec(\theta) = \frac{85}{13} }$

As We Know That,

$\implies \boxed {\sin ^{2} (\theta) + { \cos }^{2}(\theta) = 1}$

$\implies { {( \frac{13}{85} )}^{2} + { \cos }^{2}(\theta) = 1}$

$\implies { { \frac{169}{7225} } + { \cos }^{2}(\theta) = 1}$

$\implies { \cos}^{2}(\theta) = 1 - \frac{169}{7225}$

$\implies { \cos}^{2}(\theta) = \frac{7225 - 169}{7225}$

$\implies { \cos }^{2}(\theta) = \frac{7056}{7225}$

$\implies { \cos}(\theta) = \sqrt{\frac{7056}{7225}}$

$\implies { \cos}(\theta) = \frac{84}{85}$

As We Know That,

• $\sec( \theta) = \frac{1}{ \sec(\theta) }$

So,

$\implies \boxed { \sec( \theta) = \frac{85}{ 84} }$

Finding The Answer,

$\implies \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) }$

$\implies \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{ \frac{85}{13} - \frac{85}{84} }{ \frac{85}{13} + \frac{85}{84} }$

$\implies \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{ \frac{7140 - 1105}{1092} }{ \frac{7140 + 1105}{1092} }$

$\implies \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{ \frac{6035}{1092} }{ \frac{8245}{1092} }$

$\implies \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{6035}{1092} \times \frac{1092}{8245}$

$\implies \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{6035}{8245}$

$\implies \red{\boxed{ \frac{ \cosec( \theta) - \sec(\theta) }{ \cosec(\theta) + \sec(\theta) } = \frac{71}{97}}}$