NCERT Mathematics · Class 10 · English · 2026-27
Introduction to Trigonometry — Exercise 8.2
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
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Question 1
Evaluate the five standard-angle expressions in Question 1.
Step-by-step solution
Substitute exact 30°,45°,60° ratios before simplifying. Work in degrees.
(i)
\[\frac{\sqrt3}{2}\frac{\sqrt3}{2}+\frac12\frac12=1\]
(ii)
\[2(1)^2+\left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt3}{2}\right)^2=2\]
(iii)
\[\frac{\sqrt2/2}{2/\sqrt3+2}=\frac{\sqrt6}{4(1+\sqrt3)}=\frac{\sqrt6(\sqrt3-1)}8\]
(iv)
\[\frac{1/2+1-2/\sqrt3}{2/\sqrt3+1/2+1}=\frac{3\sqrt3-4}{3\sqrt3+4}\]
(v)
\[\frac{5(1/2)^2+4(2/\sqrt3)^2-1}{(1/2)^2+(\sqrt3/2)^2}=\frac54+\frac{16}3-1=\frac{67}{12}\]
Resource ID: CB-10-M-CH8-EX8.2-Q1
Question 2
Choose and justify all four standard-angle options.
Step-by-step solution
(i) [2/√3]/[1+1/3]=√3/2=sin60°, option A. (ii) (1−1)/(1+1)=0, option D. (iii) At A=0°, sin(2A)=0=2sinA. The other offered angles fail by substitution; option A. (iv) [2/√3]/[1−1/3]=√3=tan60°, option C.
Resource ID: CB-10-M-CH8-EX8.2-Q2
Question 3
Find A,B when tan(A+B)=√3 and tan(A−B)=1/√3 under the stated angle restrictions.
Step-by-step solution
With 0°<A+B≤90° and positive acute A−B, the relations give A+B=60° and A−B=30°. Add: 2A=90°; subtract: 2B=30°.
Resource ID: CB-10-M-CH8-EX8.2-Q3
Question 4
Assess the five assertions about sine, cosine and cotangent.
Step-by-step solution
Use the chapter’s interval 0°≤θ≤90°. (i) False: with A=B=30°, sin60°=√3/2 but sin30°+sin30°=1. (ii) True on this interval: sine increases from 0 to 1. (iii) False: cosine decreases from 1 to 0. (iv) False: at 30°, sine=1/2 and cosine=√3/2; equality holds at 45°, not all angles. (v) True: cot0°=cos0°/sin0°=1/0 is undefined.
Resource ID: CB-10-M-CH8-EX8.2-Q4