Question

If sec 4A = cosec (A - ${10}^o$) an 4A is acute then $\angle$ A = ?

(a) ${20}^o$ (b) ${30}^o$ (c) ${40}^o$ (d) ${50}^o$

Answer 1

$sec4A=cosec(A-{10}^o)\Rightarrow sec({90}^o-({90}^o-4A\left)\right)=cosec(A-{10}^o)\Rightarrow cosec({90}^o-4A)=cosec(A-{10}^o)\Rightarrow {90}^o-4A=A-{10}^o\Rightarrow A={20}^o$

Hence, the correct choice is (a).