​$$\frac{ 3 \sin 58^\circ}{5\cos 32^\circ}+\frac{2cos62^\circ}{5\sin28^\circ}$$​ is equal to:

Correct option is D

Given:

​$$\frac{ 3 \sin 58^\circ}{5\cos 32^\circ}+\frac{2cos62^\circ}{5\sin28^\circ}$$​​

Formula Used:

​$$sin(90-\theta)=cos\theta$$​​

​$$cos(90-\theta)=sin\theta$$​​

Solution:

We can write the given expression as,

​$$\frac{ 3 \sin 58^\circ}{5\cos 32^\circ}+\frac{2cos62^\circ}{5\sin28^\circ}=\frac{ 3 \sin (90-32)^\circ}{5\cos 32^\circ}+\frac{2cos(90-28)^\circ}{5\sin28^\circ}$$​​

Putting the values as per the formula, we get

​$$=\frac{ 3 \cos 32^\circ}{5\cos 32^\circ}+\frac{2 \sin28^\circ}{5\sin28^\circ}$$​​

​$$=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1$$​​