Question

somebody please solve this question wrong answer will be reported ​

Answer 1

$LHS = \red{ \frac{sin B}{sin A } + \frac{sec A}{sinA } }\\= \frac{sin B}{sin (90- B) } + \frac{\frac{1}{cos A}}{sinA } \\= \frac{sin B}{cos B } + \frac{1}{sinA cos A }\\= tan B + \frac{ cos^{2} A + sin^{2} A}{sinA cos A }$

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/* By Trigonometric Identity */

• cos²A + sin²A = 1

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$= tan B + \frac{cos^{2} A}{sinA cos A } +\frac{sin^{2} A}{sinA cos A } \\= tan B + \frac{cos A}{sinA } +\frac{sin A}{cos A } \\= tan B + \frac{cos (90-B)}{sin (90-B) } + tan A\\ = tan B + \frac{sin B}{cos B } + tan A\\ = tan B + tan B + tan A \\= 2tan B + tanA \\= RHS$

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