Question 126

What is the value of $$\frac{(cot\ \theta+cosec\ \theta-1)}{(cot\ \theta-cosec\ \theta+1)}$$?

Solution

ExpressionÂ :Â $$\frac{(cot\ \theta+cosec\ \theta-1)}{(cot\ \theta-cosec\ \theta+1)}$$

=Â $$(\frac{cos\theta}{sin\theta}+\frac{1}{sin\theta}-1)\div(\frac{cos\theta}{sin\theta}-\frac{1}{sin\theta}+1)$$

=Â $$(\frac{cos\theta-sin\theta+1}{sin\theta})\div(\frac{cos\theta+sin\theta-1}{sin\theta})$$

Rationalizing the denominator, we get :

= $$\frac{cos\theta-(sin\theta-1)}{cos\theta+(sin\theta-1)}\times\frac{cos\theta-(sin\theta-1)}{cos\theta-(sin\theta-1)}$$

=Â $$\frac{[cos\theta-(sin\theta-1)]^2}{cos^2\theta-(sin\theta-1)^2}$$

=Â $$\frac{cos^2\theta+(sin\theta-1)^2-2cos\theta(sin\theta-1)}{cos^2\theta-sin^2\theta-1+2sin\theta}$$

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

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

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

=Â $$\frac{(1-sin\theta)+cos\theta(1-sin\theta)}{sin\theta(1-sin\theta)}$$

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

= $$cot\theta+cosec\theta$$

=> Ans - (A)