Question

Prove that \[-1+\dfrac{\sin A\sin \left( 90-A \right)}{\cot \left( 90-A \right)}=-{{\sin }^{2}}A\].

Answer 1

Hint: We start solving the problem by recalling the trigonometric ratios $\sin \left( 90-A \right)=\cos A$ and $\cot \left( 90-A \right)=\tan A$. We substitute these results in the L.H.S of the given result and use the fact that $\tan A=\dfrac{1}{\cot A}$. We then use $\cot A=\dfrac{\sin A}{\cos A}$ and make necessary calculations in L.H.S. We then use the trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ to get the required answer.

Complete step by step answer:
We have been given a trigonometric function to solve. We need to prove that LHS = \[-{{\sin }^{2}}A\].
We have been given,
LHS = \[-1+\dfrac{\sin A\sin \left( 90-A \right)}{\cot \left( 90-A \right)}\]- (1)
We know the trigonometric ratios that are,\[\sin \left( 90-\theta \right)=\cos \theta \], \[\cot \left( 90-\theta \right)=\tan \theta \].
Using these trigonometric ratios we get \[\sin \left( 90-A \right)=\cos A\] and \[\cot \left( 90-A \right)=\tan A\].
Thus substitute these values in (1).
LHS = \[-1+\dfrac{\sin A\sin \left( 90-A \right)}{\cot \left( 90-A \right)}=-1+\dfrac{\sin A\cos A}{\tan A}\].
We know that \[\tan \theta =\dfrac{1}{\cot \theta }\].
\[\therefore \dfrac{1}{\tan A}=\cot A\].
\[\therefore \] LHS = \[-1+\sin A\cos A\cot A\] - (2)
We know that \[\cot A=\dfrac{\cos A}{\sin A}\], now put the same in (2).
LHS = \[-1+\sin A\cos A\dfrac{\cos A}{\sin A}\].
By simplifying the above, we get
LHS = \[-1+{{\cos }^{2}}A\].
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].
From this we can say that \[\Rightarrow -1+{{\cos }^{2}}A=-{{\sin }^{2}}A\].
\[\therefore \] LHS = \[-1+{{\cos }^{2}}A=-{{\sin }^{2}}A\].
We got LHS = RHS = \[-{{\sin }^{2}}A\].
Hence proved

Note: The cotangent of (90 - x) should be the same as the tangent of x. This states that the graph of a tangent function will be same as shifting of the graph of the cotangent function which is \[{{90}^{\circ }}\] to the right which means that, \[\cot \left( 90-x \right)=\tan x\]. In the problem, we should see that $-1$ doesn’t have a denominator. If we have assumed that $-1$ has a denominator, then the result varies with the given one.