- GST
- Banking
- Shares and Dividends
- Quadratic Equations in One Variable
- Factorization
- Ratio and Proportion
- Matrices
- Arithmetic and Geometric Progression
- Reflection
- Section Formula
- Equation of Straight Line
- Similarity
- Locus
- Circles
- Constructions
- Mensuration
- Trigonometric Identities
- Trigonometric Tables
- Heights and Distances
ML Aggarwal Solutions Class 10 Mathematics Solutions for Trigonometric Identities Exercise 18 in Chapter 18 - Trigonometric Identities
Prev
Prove that following:
sin (90o – θ) cos (90o – θ) = tan θ/ (1 + tan2 θ)
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}
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
