Question

prove that cos theta upon 1 minus 10 theta + sin theta upon 1 minus cot theta is equal to cos theta + sin​

Answer 1

$\text{consider}$

$=\displaystyle{\frac{cos\theta}{1-tan\theta}+\frac{sin\theta}{1-cot\theta}$

$=\displaystyle{\frac{cos^2\theta}{1-\frac{sin\theta}{cos\theta}}+\frac{sin\theta}{1-\frac{cos\theta}{sin\theta}}}$

$=\displaystyle{\frac{cos^2\theta}{cos\theta-sin\theta}+\frac{sin^2\theta}{sin\theta-cos\theta}}$

$=\displaystyle{\frac{cos^2\theta}{cos\theta-sin\theta}-\frac{sin^2\theta}{cos\theta-sin\theta}$

$=\displaystyle{\frac{cos^2\theta-sin^2\theta}{cos\theta-sin\theta}}$

$\text{Using}$

$\boxed{\bf\;a^2-b^2=(a-b)(a+b)}$

$=\displaystyle{\frac{(cos\theta-sin\theta)(cos\theta+sin\theta)}{cos\theta-sin\theta}}$

$=\displaystyle{cos\theta+sin\theta}$

$\therefore\boxed{\bf\frac{cos\theta}{1-tan\theta}+\frac{sin\theta}{1-cot\theta}=cos\theta+sin\theta}$