QUESTION

Prove the given trigonometric equation$\dfrac{{\cos \left( {90 - A} \right).\sin \left( {90 - A} \right)}}{{\tan \left( {90 - A} \right)}} = {\sin ^2}A$

Hint – In this question use trigonometric identities which are cos (90 – A) = sin (A), sin (90 – A) = cos (A) and tan (90 – A) = cot (A). Then using the basic trigonometric conversion of tan (A) in terms of cos (A) and sin(A) we can get the answer.

Given trigonometric equation is
$\dfrac{{\cos \left( {90 - A} \right).\sin \left( {90 - A} \right)}}{{\tan \left( {90 - A} \right)}} = {\sin ^2}A$
Proof –
Consider L.H.S
$\Rightarrow \dfrac{{\cos \left( {90 - A} \right).\sin \left( {90 - A} \right)}}{{\tan \left( {90 - A} \right)}}$
Now as we know cos (90 – A) = sin (A), sin (90 – A) = cos (A) and tan (90 – A) = cot (A), so substitute these values in above equation we have,
$\Rightarrow \dfrac{{\sin A.\cos A}}{{\cot A}}$
Now as we know that cot is the ratio of cosine to sine so use this property in above equation we have,
$\Rightarrow \dfrac{{\sin A.\cos A}}{{\dfrac{{\cos A}}{{\sin A}}}}$
Now simplify the above equation we have,
$\Rightarrow \dfrac{{\sin A.\cos A.\sin A}}{{\cos A}}$
Now as we see that cos (A) is cancel out from numerator and denominator so we have,
$\Rightarrow {\sin ^2}A$
= R.H.S
Hence Proved.

Note – Problems of this kind are solemnly based upon trigonometric identities, it is advised to remember all trigonometric based identity. It’s difficult to grasp them all but practice always helps. Some of the important identities include sin (A+B) = sin A cos B + cos A sin B. The key point in this question is the breakdown of tan A into sin A and cos A, similarly sec A can also be written and it will simply be the converse of tan A.