Question

If $\sec \beta \sec \gamma +\tan \beta \tan \gamma =sec\alpha$. Find $\sec \beta$

• A. $\frac{\sec \alpha \cos \gamma }{1+\sin \beta \sin \gamma }$
• B. $\frac{\sec \alpha \cos \gamma }{2+\sin \beta \sin \alpha }$
• C. $\frac{\cos \alpha \cos \gamma }{1+\sin \beta \sin \gamma }$
• D. $\frac{\cos \beta .\cos \gamma -\cos \alpha .\sin \beta .sin\gamma }{\cos \alpha .\cos \beta }$

Answer 1

Answer: (D)

$\frac{\cos \beta .\cos \gamma -\cos \alpha .\sin \beta .sin\gamma }{\cos \alpha .\cos \beta }$

$\sec \beta \sec \gamma +\tan \beta \tan \gamma =\sec \alpha$

$\Rightarrow \sec \beta =\frac{\sec \alpha -\tan \beta \tan \gamma }{\sec \gamma }$

$\Rightarrow \sec \beta =\frac{\frac{1}{\cos \alpha }-\frac{\sin \beta }{\cos \beta }\times \frac{\sin \gamma }{\cos \gamma }}{\frac{1}{\cos \gamma }}$

$\Rightarrow \sec \beta =\frac{\cos \beta \cos \gamma -\cos \alpha \sin \beta \sin \gamma }{\cos \alpha \cos \beta }$