Question

please solve this trigonometry <br />to prove ques​

Answer 1

$\huge \boxed{\red{ \mathfrak{solution}}}$

To prove :

$\bold{ \sqrt{ \frac{cosec \: x - 1}{cosec \: x + 1} } = \frac{1}{sec \: x + tan \: x}}$

LHS

$\implies \bold{ \sqrt{ \frac{cosec \: x - 1}{cosec \: x + 1} } }$

Rationalise the denominator :

$\implies \bold{ \sqrt{ \frac{cosec \: x - 1}{cosec \:x + 1} \times \frac{cosec \: x - 1}{cosec \: x - 1} } }\\ \\ \implies \bold{ \sqrt{ \frac{( {cosec \: x - 1)}^{2} }{ {cosec}^{2}x - {1}^{2} } } } \\ \\ \implies \bold{ \sqrt{ \frac{( {cosec \: x - 1)}^{2} }{ {cot}^{2} x} } } \\ \\ \implies \bold{ \frac{cosec \: x - 1}{cot \: x} } \\ \\ \implies \bold{ \frac{cosec \: x}{cot \: x} - \frac{1}{cotx} } \\ \\ \implies \bold{ \frac{ \frac{1}{sinx} }{ \frac{cosx}{sinx} } - \: \frac{1}{ \frac{cosx}{sinx} } } \\ \\ \implies \bold{ \frac{ \frac{1}{ \cancel{sinx}} }{ \frac{cosx}{ \cancel{sinx}} } - \frac{sinx}{cosx} } \\ \\ \implies \bold{ \frac{1}{cosx} - \frac{sinx}{cosx} } \\ \\ \implies \bold{secx \: - tanx}$

Now again rationalise the numerator :

=> secx -tanx ×(secx+tanx)/(secx+tanx)

=> sec^2x-tan^2x/ secx + tanx

=> 1/ secx + tanx

proved .

Formula :

• cosec^2x-cot^2x=1

• sec^2x-tan^2x=1