Question

It $\sin \theta +\cos \theta =a$ and $\sec \theta +\text{cosec}\theta =b$, find the value of $b(a^2-1)$.

• A. $2a$
• B. $3a$
• C. $0$
• D. $2ab$

Answer 1

Answer: (A)

$2a$

Given, $\sin \theta +\cos \theta =a,\sec \theta +\text{cosec}\theta =b$

Therefore, $a^2=1+2\sin \theta .\cos \theta$

$\Rightarrow a^2-1=2\sin \theta .\cos \theta$

$\Rightarrow b=\frac{1}{\cos \theta }+\frac{1}{\sin \theta }=\frac{\sin \theta +\cos \theta }{\sin \theta .\cos \theta }$

Thsu $(a^2-1)b=2(\sin \theta +\cos \theta )=2a$