NCERT Trigonometry Class 10 Exercise 8.2 Solutions

What is the trigonometry ratios table?

It is a type of table by this table we can find out the value of some specific angles like 0º,30º,45°,60°and 90º. So it is essential that you know how to make this table, here I am going to explain this table in a very easy manner here I have also solved ncert class 10 trigonometry exercise 8.2 in a very simple way.

	0º (√o)	30º(√1/4)	45º(√1/2)	60º(√3/4)	90º√1
sinθ	0	1/2	1/√2	√3/2	1	square root of numbers
cosθ	1	√3/2	1/√2	1/2	0	reverse order of sinθ
tanθ	0	1/√3	1	√3	∞	sinθ/cosθ
cotθ	∞	√3	1	1/√3	0	reverse order of tanθ
secθ	1	2/√3	√2/1	2	∞	reciprocal of cosθ
cosecθ	∞	2	√2/1	2/√3	1	reverse order of cosθ

Q.1) Evaluate the following :

(i) sin60ºcos30º+sin30ºcos60º

= √3/2×√3/2+ 1/2 ×1/2 [put the value of all trigonometric ratios with the help of the table]

= 3/4 + 1/4

= 4/4

= 1

(ii) 2tan²45º + cos²30º – sin²60º

solution:

2tan²45º + cos²30º – sin²60º

= 2(1)²+(√3/2)²-(√3/2)²

= 2×1 + 3/4 – 3/4

= 2

(iii) cos45º/sec30º+cosec30º

solution:

cos45º/sec30º+cosec30º

= (1/√2)/(  2/√3)+2

= (1/√2)/[(2+2√3)/√3]

= (1/√2)×√3/2+2√3

= √3/(2+2√3)

= √3/2(1+√3)

= √3/2(√3+1)

= √3/2√2(√3+1)  × √2/√2     {Rationalization of root 2}

= √6/2×2(√3+1)

= √6/4(√3+1)  ×(√3-1)/(√3-1)     {Rationalization of  √3+1 }

=√6(√3-1)/4(3-1)     [(a+b)(a-b) = a² – b²]

= √18-√6/4×2

= {√(2×3×3) – √6}/8

= {3√2 – √6}/8

(iv) sin30º+tan45º- cosec60º /  sec30º+cos60º + cot45º

solution:

sin30º+tan45º- cosec60º /  sec30º+cos60º + cot45º

= ( 1/2)+1- 2/√3   /  [2/√3 + 1/2 + 1]

= [(√3 +2√3 -4)/2√3 ] / [(4+√3 +2√3)/2√3  ]

= [(3√3 -4)/2√3 ] / [(4+3√3 )/2√3 ]

= 3√3 -4/ 4+3√3

= 3√3 -4/ 3√3 +4     [Rationalise the denominator]

= 3√3 -4 / 3√3 +4 × 3√3 -4 / 3√3 -4          [ (a+b)(a-b)=a²-b² ]

= ( 3√3 -4 )² / (3√3)²-(4)²           

= (3√3)² -2×3√3×4 +(4)² / 9×3 –  16

= [9×3 -24√3+16] / 27-16

= [27-24√3+16] / 11

= 43-24√3 / 11

(v) 5cos²60º + 4sec²30º- tan²45º   /   sin²30º + cos²30º

solution:

5cos²60º + 4sec²30º- tan²45º   /   sin²30º + cos²30º

= 5(1/2)² + 4( 2/√3)²– ( 1)²   /   (1/2)² + (√3/2)²

= [5×1/4 + 4×4/3 – 1 ]   /  (1/4 + 3/4)

= [5/4 + 16/3 – 1]    /   (4/4)

= [(15+64-12)/12 ] /   (1)

= [(79-12)/12]

= [(67/12)]

= 67/12

Q2.) Choose the correct option and justify your choice :

(i) 2 tan30º / 1 + tan²30 = 

(A) sin60º        (B) 1                (C) sin45º        (D) sin30º

Explaination:

2 tan30º / 1 + tan²30 

= 2×(1/√3) / 1+ (1/√3)²

= 2/√3  /  1+1/3

= 2/√3 ÷ 4/3

= (2 /√3) ×3/4

= 2×3/√3 ×4

= √3×√3 / 2√3

= √3/2

= sin60º

Ans ( A)

(ii) 1-tan²45º /1+tan²45º

Solution:

1-tan²45º /1+tan²45º

= 1-(1)² / 1+(1)²

=1-1/1+1

=0/2

=0

Ans(D) 0

(iii) sin 2A = 2 sinA is true when A =

(A) 0°            (B) 30º            (C) 45°                   (D) 60°

Explaination:

put A = 0º

sin 2×0º = 2 sin0º

sin 0º = 2sin0º

=> 0 = 2×0

=> 0 = 0

=> LHS = RHS

Ans (A) 0º

(iv) 2 tan30º / 1-tan²30º =

(A) cos 60º                   (B) sin 60º      (C) tan60º                (D) sin 30º

Explaination:

2 tan30º / 1-tan²30º

= 2×(1/√3) / 1- (1/√3)²

= (2/√3 ) /  1 – (1/3)

= (2/√3 ) / (3- 1)/3

= (2/√3 )/(2/3)

= (2/√3) ×( 3/2)

= 2×3/2√3

= 3/√3

= √3

= tan60º

Ans(C)  tan60º

Q3.) If tan ( A+B )= √3 and tan(A – B) =1/√3; 0º< A+B≤90º; A>B , find A and B.

Given:

tan ( A+B )= √3;

tan(A – B) =1/√3;

0º< A+B≤90º; A>B

To Find:

A=?

B=?

Solution:

tan ( A+B )= √3

=>tan ( A+B )= tan60º

=> A+B=60º  …………….(i)

tan(A – B) =1/√3

=> tan(A – B) = tan 30º

=>A – B = 30º     ……………….(ii)

Adding Equation(i) and Equation (ii)

A+B + A-B=60º +30º

=> 2A = 90º

=> A = 90º/2

=> A = 45  [put into Equation (i)]

45 + B = 60º

=> B = 60º-45º

=> B = 15º

Justification

0º< A+B ≤90º

0º< 45º+15º ≤90º

0º< 60º ≤ 90º  and

A > B

=> 45º > 15º

Hence

A = 45º And B = 15º