Question

Prove that $\sin \left( {90 - \theta } \right)\,\,\cos \left( {90 - \theta } \right) = \dfrac{{\tan \theta }}{{1 + {{\cot }^2}\left( {90 - \theta } \right)}}$ ?

Answer 1

Here in this question, we have to prove the given trigonometric function by showing the left hand side is equal to the right hand side (i.e., $L.H.S = R.H.S$). To solve this, we have to consider L.H.S and R.H.S separately and simplify by using a definition and complementary angles of trigonometric ratios and by trigonometric identities to get the required solution.

Complete step by step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
Prove that
$\sin \left( {90 - \theta } \right)\,\,\cos \left( {90 - \theta } \right) = \dfrac{{\tan \theta }}{{1 + {{\cot }^2}\left( {90 - \theta } \right)}}$ --------(1)
Consider Left hand side of equation (1) (L.H.S)
$\Rightarrow \,\,\,\sin \left( {90 - \theta } \right)\,\,\cos \left( {90 - \theta } \right)$ ----(2)

Let us by the complementary angles of trigonometric ratios:
The angle can be written as
$\sin \left( {90 - \theta } \right) = \cos \theta$
$\Rightarrow \cos \left( {90 - \theta } \right) = \sin \theta$
On substituting in equation (2), we have
$\therefore \,\,\cos \theta \,\,\sin \theta$-----(3)
Consider Right hand side of equation (1) (R.H.S)
$\Rightarrow \,\,\dfrac{{\tan \theta }}{{1 + {{\cot }^2}\left( {90 - \theta } \right)}}$ -------(4)
By complementary angle ${\cot ^2}\left( {90 - \theta } \right) = {\tan ^2}\theta$, then on substituting we have
$\Rightarrow \,\,\dfrac{{\tan \theta }}{{1 + {{\tan }^2}\theta }}$

As we know the trigonometric identities: $1 + {\tan ^2}\theta = {\sec ^2}\theta$ , then equation (4) becomes
$\Rightarrow \,\,\dfrac{{\tan \theta }}{{{{\sec }^2}\theta }}$
As by the definition of trigonometric ratios: $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$.
On substituting, we have
$\Rightarrow \,\,\dfrac{{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}}{{\left( {\dfrac{1}{{{{\cos }^2}\theta }}} \right)}}$
Or it can be written as
$\Rightarrow \,\,\dfrac{{\sin \theta }}{{\cos \theta }} \times \left( {{{\cos }^2}\theta } \right)$
On simplification, we get
$\therefore \,\,\sin \theta \,\,\cos \theta \,\,$ ------(5)
From equation (3) and (5)
$\sin \left( {90 - \theta } \right)\,\,\cos \left( {90 - \theta } \right) = \dfrac{{\tan \theta }}{{1 + {{\cot }^2}\left( {90 - \theta } \right)}}$
Hence, proved.

Note: When solving the trigonometry-based questions, we have to know the definitions of six trigonometric ratios. Remember, when the sum of two angles is ${90^ \circ }$, then the angles are known as complementary angles at that time the ratios will change like $\sin \leftrightarrow \cos$, $\sec \leftrightarrow cosec$ and $\tan \leftrightarrow \cot$ then should know the value of standard angles and basic three trigonometric identities.