Question

If A, B, C, are acute angles and $\sin A = \cos A$ and $\sin B\cos C + \cos B\sin C = \sin A$ then tan A is equal to ?
(A) $\tan B + \tan C$
(B) $2(\tan B + \tan C)$
(C) $\tan B + 2\tan C$
(D) $2\tan B + \tan C$

Answer 1

Hint: 1. Trigonometric ratios of $(90^\circ - \theta )$
$\sin (90^\circ - \theta ) = \cos \theta $
$\cos (90^\circ - \theta ) = \sin \theta $
$\tan (90^\circ - \theta ) = \cot \theta $
$\sec (90^\circ - \theta ) = \csc \theta $
$\csc (90^\circ - \theta ) = \sec \theta $
$\cot (90^\circ - \theta ) = \tan \theta $
2. Addition and subtraction of angles in sin
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\sin (A - B) = \sin A\cos B - \cos A\sin B$

Complete step-by-step answer:
It is given that
1. $\sin A = \cos B$
2. $\sin B\cos C + \cos B\sin C = sinA$
Since $\sin A = \cos B$
Then $\sin A = \sin \left( {\dfrac{\pi }{2} - B} \right)$ $\left( {\because \sin (90^\circ - \theta ) = \cos \theta } \right)$
So, $A = \dfrac{\pi }{2} - B$
Or $A + B = \dfrac{\pi }{2}$ .….(1)
Now, we know that
$\sin (a + b) = \sin a \times \cos b + \cos a \times \sin b$
So, by using this formula we can say that
The given equation
$\sin B\cos C + \cos B\sin C = \sin (B + C) = \sin A$
Or, $\sin (B + C) = \sin A$
Taking tangent both side
$\tan (B + C) = \tan A$
or $\tan A = \tan (B + C)$
$\tan A = \dfrac{{\tan B + \tan C}}{{1 - \tan B\tan C}}$
$\tan A - \tan A\tan B\tan C = \tan B + \tan C$
From equation (1) $\left( {A = \dfrac{\pi }{2} - B} \right)$
$\tan A - \tan \left( {\dfrac{\pi }{2} - B} \right)\tan B\tan C = \tan B + \tan C$
$\tan A - \cot B.\tan B.\tan C = \tan B + \tan C$
$\tan A - \tan C = \tan B + \tan C$
Or, $\tan A = \tan B + 2\tan C$
Therefore, option C is correct option i.e., $\tan A = \tan B + 2\tan C$
So, the correct answer is “Option C”.

Note: In solution part we cut off tan B from cot B because they both are inverse of each other
Or, $\tan B = \dfrac{1}{{\cot B}}$
Similarly try to remember all the reciprocal relations and other relations between trigonometric functions.