# Complete NCERT solutions for class 10 Maths Chapter 8 Introduction To Trigonometry

Written by Team Trustudies
Updated at 2021-02-16

Solution for Exercise -8.1

Q1. In , right angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A
(ii) sin C, cos C

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given, in ,
AB = 24 cm, BC = 7 cm and right angled at B

By using the Pythagoras theorem, we have

=>
=>

=> ? cm

(i) ???? [? ]

???? [? ]

(ii)

Q2. In Fig. find tan P – cot R.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

By using the Pythagoras theorem, we have

=>

=>

=>

?

Hence,

Q3. If , calculate cos A and tan A.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given,

Let BC = 3k and AC = 4k.

Then,

?

Q4. Given 15 cot A = 8, find sin A and sec A.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given,

=>

Let AB = 8k and BC = 15k.

Then, by Pythagoras theorem

?

and,

Q5. Given , calculate all other trigonometric ratios.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given,

Let AC = 13k and AB = 12k.

Then by Pythagoras theorem,

?

Q6. If and are acute angles such that cos A = cos B, then show that .

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

In ,

and,

But given that, cos A = cos B

=>
=>

? [? Angles opposite to equal sides are equal]

Q7 If , evaluate :

(i) $\frac{\left(1+sin?\right)\left(1?sin?\right)}{\left(1+cos?\right)\left(1?cos?\right)}$

(ii) $co{t}^{2}?$

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given,

Let AB = 7k and BC = 8k.
Then by Pythagoras theorem,

?

and,

(i) ?

= $\frac{1?\frac{64}{113}}{1?\frac{49}{113}}$

= $\frac{113?64}{113?49}$

= $\frac{49}{64}$

(ii)

Q8. If , check whether or not.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given,

Let AB = 4k and BC = 3k

Then by Pythagoras theorem,

?

and,

Now, L.H.S. =

R.H.S. =

? L.H.S = R.H.S.

?

Q9. In right angled at B, if , find the value of

(i)
(ii) $cosAcosC?sinAsinC$

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given,

Let and .

Then by Pythagoras theorem,

?

and,

(i)

(ii)

Q10. In , right angled at Q, cm and PQ = 5 cm. Determine the values of sin P, cos P , and tan P.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Given, PR + QR = 25 cm and PQ = 5cm

Let QR = x cm
Thus, PR = (25 – x) cm

Then by Pythagoras theorem,

=>

? RQ = 12 cm

=> RP = (25 – 12) = 13 cm

Now,

and

Q11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.

(ii) for some value of angle A .

(iii) $cosA$ is the abbreviation used for the cosecant of angle A.

(iv) $cotA$ is the product of cot and A.

(v) for some angle $?$.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

(i) False because sides of a right triangle may have any length, so tan A may have any value.

(ii) True as sec A is always greater than 1.

As, and Hypotenuse is the largest side ? sec A is always greater than 1

(iii) False as cos A is the abbreviation used for cosine A.

(iv) False as cot A is not the product of 'cot' and A. 'cot' separated from A has no meaning.

(v) False as sin $?$ cannot be > 1

Solution for Exercise -8.2

Q1. Evaluate :
(i) $sin60°cos30°+sin30°cos60°$

(ii) $2ta{n}^{2}45°+co{s}^{2}30°?si{n}^{2}60°$

(iii) $\frac{cos45°}{sec30°+cosec30°}$

(iv) $\frac{sin30°+tan45°?cosec60°}{sec30°+cos60°+cot45°}$

(v) $\frac{5co{s}^{2}60°+4se{c}^{2}30°?ta{n}^{2}45°}{si{n}^{2}30°+co{s}^{2}30°}$

NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

(i)

(ii)

(iii)

(iv) $\frac{sin30°+tan45°?cosec60°}{sec30°+cos60°+cot45°}$

(v) $\frac{5co{s}^{2}60°+4se{c}^{2}30°?ta{n}^{2}45°}{si{n}^{2}30°+co{s}^{2}30°}$

Q2. Choose the correct option and justify :

(i) $\frac{2tan30°}{1+ta{n}^{2}30°}$ =

(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) $\frac{1?ta{n}^{2}45°}{1+ta{n}^{2}45°}$ =

(A) tan 90°
(B) 1
(C) sin 45 °
(D) 0

(iii)