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

Question 29 Trigonometric Identities Exercise 18

\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\cot ^{2} A}{1+\cot ^{2} A}=1

In LHS, we need to convert cot2 A in terms of tan2 A for second term.

\begin{aligned} &\text { L.H.S. }=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\frac{1}{\tan ^{2} A}}{1+\frac{1}{\tan ^{2} A}}\\ &=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\frac{1}{\tan ^{2} A}}{\frac{\tan ^{2} A+1}{\tan ^{2} A}}\\ &=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{\tan ^{2} A}{\tan ^{2} A\left(\tan ^{2} A+1\right)}\\ &=\frac{\tan ^{2} A}{1+\tan ^{2} A}+\frac{1}{1+\tan ^{2} A}\\ &=\frac{1+\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}=1=\mathrm{R} . \mathrm{H} . \mathrm{S} . \end{aligned}

Related Questions
Exercises

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!