Question

cos theta by 1 minus 10 theta + sin theta by 1 minus cot theta is equal to sin theta + cos theta​

Answer 1

Answer:

Step-by-step explanation:

Given that:

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

Solution:

As we know that

$tan\theta=\frac{sin\theta}{cos\theta}\\\\cot\theta=\frac{cos\theta}{sin\theta}$

Put these value in the LHS

$\frac{cos\theta}{1-\frac{sin\theta}{cos\theta}}+\frac{sin\theta}{1-\frac{cos\theta}{sin\theta}}$

Take LCM in the denominator

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

Take - sign common from the denominator of second term

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

Take LCM

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

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

Hence prove.

Answer 2

Answer:

$follow \: the \: attchment$