NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1#
Exercise 8.1 introduces trigonometric ratios — sine, cosine, tangent, cosecant, secant, and cotangent — defined with respect to the sides of a right-angled triangle. Students learn to find these ratios from given information.
Trigonometric Ratios in right △ABC (right angle at C): $\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AC}$, $\cos A = \dfrac{AB}{AC}$, $\tan A = \dfrac{BC}{AB}$ $\csc A = \dfrac{1}{\sin A}$, $\sec A = \dfrac{1}{\cos A}$, $\cot A = \dfrac{1}{\tan A}$
1. In △ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine: #
Setup
By Pythagoras Theorem:
$$AC = \sqrt{AB^2 + BC^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm}$$(i) sin A, cos A
(ii) sin C, cos C
2. In the figure, find tan P – cot R. #
Answer
From the figure: In △PQR right-angled at Q, $PQ = 12$ cm, $PR = 13$ cm.
$$QR = \sqrt{PR^2 - PQ^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}$$$$\tan P = \frac{QR}{PQ} = \frac{5}{12}$$$$\cot R = \frac{QR}{PQ} = \frac{5}{12}$$$$\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = \mathbf{0}$$3. If sin A = $\dfrac{3}{4}$, calculate cos A and tan A. #
Answer
$\sin A = \dfrac{3}{4}$, so $\cos^2 A = 1 - \sin^2 A = 1 - \dfrac{9}{16} = \dfrac{7}{16}$
$$\cos A = \frac{\sqrt{7}}{4}$$$$\tan A = \frac{\sin A}{\cos A} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$$4. Given 15 cot A = 8, find sin A and sec A. #
Answer
$15 \cot A = 8 \Rightarrow \cot A = \dfrac{8}{15}$
So $\tan A = \dfrac{15}{8}$
In right triangle: opposite = 15, adjacent = 8, hypotenuse $= \sqrt{225+64} = \sqrt{289} = 17$
$$\sin A = \frac{15}{17}$$$$\sec A = \frac{17}{8}$$5. Given sec θ = $\dfrac{13}{12}$, calculate all other trigonometric ratios. #
Answer
$\sec\theta = \dfrac{13}{12}$, so hypotenuse = 13, adjacent = 12.
Opposite $= \sqrt{169-144} = \sqrt{25} = 5$
$$\sin\theta = \frac{5}{13}, \quad \cos\theta = \frac{12}{13}, \quad \tan\theta = \frac{5}{12}$$$$\csc\theta = \frac{13}{5}, \quad \sec\theta = \frac{13}{12}, \quad \cot\theta = \frac{12}{5}$$6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B. #
Proof
In a right triangle, $\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}$.
If $\cos A = \cos B$, then the ratios of adjacent side to hypotenuse are equal for both angles.
More formally: In △ABC right-angled at C, $\cos A = \dfrac{AC}{AB}$ and $\cos B = \dfrac{BC}{AB}$
If $\cos A = \cos B$:
$$\frac{AC}{AB} = \frac{BC}{AB} \Rightarrow AC = BC$$In △ABC, if $AC = BC$, then the angles opposite equal sides are equal:
$$\angle A = \angle B$$Hence proved. $\blacksquare$
7. If cot θ = $\dfrac{7}{8}$, evaluate: #
(i) $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$\cot\theta = \dfrac{7}{8}$: adjacent = 7, opposite = 8, hypotenuse $= \sqrt{49+64} = \sqrt{113}$
$$\sin\theta = \frac{8}{\sqrt{113}}, \quad \cos\theta = \frac{7}{\sqrt{113}}$$$$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} = \frac{1-\sin^2\theta}{1-\cos^2\theta} = \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \frac{49}{64}$$(ii) $\cot^2\theta$
8. If 3 cot A = 4, check whether $\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A$ or not. #
Answer
$3\cot A = 4 \Rightarrow \cot A = \dfrac{4}{3} \Rightarrow \tan A = \dfrac{3}{4}$
Adjacent = 4, opposite = 3, hypotenuse = 5. $\sin A = \dfrac{3}{5}$, $\cos A = \dfrac{4}{5}$
LHS:
$$\frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{25}$$RHS:
$$\cos^2 A - \sin^2 A = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$Since LHS = RHS = $\dfrac{7}{25}$, the identity holds. ✓
9. In triangle ABC, right-angled at B, if tan A = $\dfrac{1}{\sqrt{3}}$, find the value of: #
(i) sin A cos C + cos A sin C
$\tan A = \dfrac{1}{\sqrt{3}} \Rightarrow \angle A = 30°$, so $\angle C = 60°$.
$$\sin A = \frac{1}{2}, \cos A = \frac{\sqrt{3}}{2}, \sin C = \frac{\sqrt{3}}{2}, \cos C = \frac{1}{2}$$$$\sin A \cos C + \cos A \sin C = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = \mathbf{1}$$(This is $\sin(A+C) = \sin 90° = 1$)
(ii) cos A cos C – sin A sin C
(This is $\cos(A+C) = \cos 90° = 0$)
10. In △PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. #
Answer
Let $QR = x$. Then $PR = 25 - x$.
By Pythagoras Theorem:
$$PR^2 = PQ^2 + QR^2$$$$(25-x)^2 = 25 + x^2$$$$625 - 50x + x^2 = 25 + x^2$$$$600 = 50x$$$$x = 12$$So $QR = 12$ cm, $PR = 13$ cm.
$$\sin P = \frac{QR}{PR} = \frac{12}{13}$$$$\cos P = \frac{PQ}{PR} = \frac{5}{13}$$$$\tan P = \frac{QR}{PQ} = \frac{12}{5}$$11. State whether the following are true or false. Justify your answer. #
(i) The value of tan A is always less than 1.
False.
$\tan A$ can be any positive value. For example, $\tan 60° = \sqrt{3} > 1$.
$\tan A < 1$ only when $A < 45°$.
(ii) sec A = $\dfrac{12}{5}$ for some value of angle A.
True.
$\sec A = \dfrac{12}{5} > 1$, which is valid since $\sec A \geq 1$ for all acute angles A. (Here $A = \cos^{-1}\dfrac{5}{12}$)
(iii) cos A is the abbreviation used for the cosecant of angle A.
False.
$\cos A$ is the abbreviation for cosine of angle A. The cosecant of angle A is abbreviated as $\csc A$ (or $\text{cosec}\, A$).
(iv) cot A is the product of cot and A.
False.
$\cot A$ is the cotangent of angle A. It is not a product — “cot” has no meaning without the angle A.
(v) sin θ = $\dfrac{4}{3}$ for some angle θ.
False.
The range of $\sin\theta$ is $[-1, 1]$. Since $\dfrac{4}{3} > 1$, there is no angle for which $\sin\theta = \dfrac{4}{3}$.