Question

In a triangle, if cosec (a + b) = sec 3(a + b), then the value of tan 10(a + b) is

• A.
• B. 1
• C. 0
• D.

Answer 1

The correct option is B

1

$\begin{array}{c}cosec\left(a+b\right)=\sec 3\left(a+b\right)\\ \Rightarrow sin\left(a+b\right)=cos3\left(a+b\right)\\ \Rightarrow sin\left(a+b\right)=sin\left[90-3\left(a+b\right)\right]Usingcosx=sin\left(90-x\right)\\ \Rightarrow a+b=90-3a-3b\\ \Rightarrow 4\left(a+b\right)=90\\ \Rightarrow a+b=90/4\end{array}$

Now substitute the value of a + b in tan10 (a + b)

$\therefore$tan 10*(90/4) = tan 225${}^{\circ }$ = tan 45${}^{\circ }$ = 1.