# ML Aggarwal Solutions Class 10 Mathematics Solutions for Trigonometric Identities Exercise 18 in Chapter 18 - Trigonometric Identities

Question 15 Trigonometric Identities Exercise 18

Prove that following:

sin (90o – θ) cos (90o – θ) = tan θ/ (1 + tan2 θ)

Answer:

Here we can apply the identity sin2 θ + cos2 θ =1

\begin{aligned} &\sin \left(90^{\circ}-\theta\right) \cos \left(90^{\circ}-\theta\right)=\frac{\tan \theta}{1+\tan ^{2} \theta}\\ &\text { R.H.S. }=\frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}\\ &=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}}\\ &=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos ^{2} \theta}}=\frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta=\sin \theta \cos \theta\\ &\therefore \text { L.H.S. }=\text { R.H.S. } \end{aligned}

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