Question 69

If $$\frac{cos\theta}{1-sin\theta}+\frac{cos\theta}{1+sin\theta}=4$$ then the value of $$\theta(0<\theta<90^{\circ})$$ is

Solution

Given : $$\frac{cos\theta}{1-sin\theta}+\frac{cos\theta}{1+sin\theta}=4$$

=> $$\frac{cos\theta(1+sin\theta)+cos\theta(1-sin\theta)}{(1-sin\theta)(1+sin\theta)}=4$$

=> $$(cos\theta+cos\theta sin\theta)+(cos\theta-cos\theta sin\theta)=4(1-sin^2\theta)$$

=> $$2cos\theta=4cos^2\theta$$

=> $$2cos\theta=1$$

=> $$cos\theta=\frac{1}{2}$$

=> $$\theta=cos^{-1}(\frac{1}{2})$$

=> $$\theta=60^\circ$$

=> Ans - (A)

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