Question

(1/secα-tanα)-1/cosα=1/cos​

Answer 1

$\textit{\large{\underline{\underline{\pink{Step by step Explanation:-}}}}}$

$\mathrm{\frac{1}{secA-tanA}-\frac{1}{cosA}=tanA}$

$\mathrm{LHS=\frac{1}{secA-tanA}-\frac{1}{cosA}}$

$\mathrm{=\frac{1}{\frac{1}{cosA}-\frac{sinA}{cosA}}-\frac{1}{cosA}}$

$\mathrm{=\frac{1}{\frac{1-sinA}{cosA}}-\frac{1}{cosA}}$

$\mathrm{=\frac{cosA}{1-sinA}-\frac{1}{cosA}}$

$\mathrm{=\frac{cos^{2}A-(1-sinA)}{cosA(1-sinA)}}$

$\mathrm{=\frac{1-sin^{2}A-1+sinA}{cosA(1-sinA)}}$

$\mathrm{=\frac{sinA-sin^{2}A}{cosA(1-sinA)}}$

$\mathrm{=\frac{sinA(1-sinA)}{cosA(1-sinA)}}$

$\mathrm{=\frac{sinA}{cosA}}$

$\mathrm{=tanA}$

❣️❣️✌️✌️Hope it helps you ❣️❣️✌️✌️