Chapter 8 · Introduction to Trigonometry and its Applications

NCERT Exemplar Class 10 Maths Exercise 8.1

All 15 questions of NCERT Exemplar Class 10 Maths Exercise 8.1 (Introduction to Trigonometry and its Applications), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   If $\cos A=\frac{4}{5}$, ​then the ​value ​of $\tan A$ is

   1. (a)$\frac{3}{5}$
   2. (b)$\frac{3}{4}$
   3. (c)$\frac{4}{3}$
   4. (d)$\frac{5}{3}$
   Show answer

   (b) $\frac{3}{4}$

   Step-by-step solution →
2. Q2
   If $\sin A=\frac{1}{2}$, ​then ​the ​value of $\cot A$ ​is

   1. (a)$\sqrt{3}$
   2. (b)$\frac{1}{\sqrt{3}}$
   3. (c)$\frac{\sqrt{3}}{2}$
   4. (d)$1$
   Show answer

   (a) $\sqrt{3}$

   Step-by-step solution →
3. Q3
   The value ​of ​the expression $[\csc (75^{\circ }+\theta )-\sec (15^{\circ }-\theta )-\tan (55^{\circ }+\theta )+\cot (35^{\circ }-\theta )]$ ​is

   1. (a)$-1$
   2. (b)$0$
   3. (c)$1$
   4. (d)$\frac{3}{2}$
   Show answer

   (b) $0$

   Step-by-step solution →
4. Q4
   Given that $\sin \theta =\frac{a}{b}$, ​then $\cos \theta$ ​is equal ​to

   1. (a)$\frac{b}{\sqrt{b^2-a^2}}$
   2. (b)$\frac{b}{a}$
   3. (c)$\frac{\sqrt{b^2-a^2}}{b}$
   4. (d)$\frac{a}{\sqrt{b^2-a^2}}$
   Show answer

   (c) $\frac{\sqrt{b^2-a^2}}{b}$

   Step-by-step solution →
5. Q5
   If $\cos (\alpha +\beta )=0$, ​then $\sin (\alpha -\beta )$ ​can ​be reduced ​to

   1. (a)$\cos \beta$
   2. (b)$\cos 2\beta$
   3. (c)$\sin \alpha$
   4. (d)$\sin 2\alpha$
   Show answer

   (b) $\cos 2\beta$

   Step-by-step solution →
6. Q6
   The ​value ​of $(\tan 1^{\circ }\tan 2^{\circ }\tan 3^{\circ }\dots \tan 89^{\circ })$ ​is

   1. (a)$0$
   2. (b)$1$
   3. (c)$2$
   4. (d)$\frac{1}{2}$
   Show answer

   (b) $1$

   Step-by-step solution →
7. Q7
   If $\cos 9\alpha =\sin \alpha$ ​and $9\alpha <90^{\circ }$, ​then ​the value of $\tan 5\alpha$ ​is

   1. (a)$\frac{1}{\sqrt{3}}$
   2. (b)$\sqrt{3}$
   3. (c)$1$
   4. (d)$0$
   Show answer

   (c) $1$

   Step-by-step solution →
8. Q8
   If $△ABC$ ​is ​right ​angled ​at $C$, ​then the value of $\cos (A+B)$ ​is

   1. (a)$0$
   2. (b)$1$
   3. (c)$\frac{1}{2}$
   4. (d)$\frac{\sqrt{3}}{2}$
   Show answer

   (a) $0$

   Step-by-step solution →
9. Q9
   If $\sin A+{\sin }^{2}A=1$, then ​the ​value of ​the ​expression $({\cos }^{2}A+{\cos }^{4}A)$ ​is

   1. (a)1
   2. (b)$\frac{1}{2}$
   3. (c)2
   4. (d)3
   Show answer

   (a) 1

   Step-by-step solution →
10. Q10
    Given ​that $\sin \alpha =\frac{1}{2}$ and $\cos \beta =\frac{1}{2}$, ​then ​the value ​of $(\alpha +\beta )$ ​is

    1. (a)$0^{\circ }$
    2. (b)$30^{\circ }$
    3. (c)$60^{\circ }$
    4. (d)$90^{\circ }$
    Show answer

    (d) $90^{\circ }$

    Step-by-step solution →
11. Q11
    The ​value of ​the ​expression $\left[\frac{{\sin }^{2}22^{\circ }+{\sin }^{2}68^{\circ }}{{\cos }^{2}22^{\circ }+{\cos }^{2}68^{\circ }}+{\sin }^{2}63^{\circ }+\cos 63^{\circ }\sin 27^{\circ }\right]$ is

    1. (a)3
    2. (b)2
    3. (c)1
    4. (d)0
    Show answer

    (b) 2

    Step-by-step solution →
12. Q12
    If $4\tan \theta =3$, ​then $\left(\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }\right)$ ​is equal to

    1. (a)$\frac{2}{3}$
    2. (b)$\frac{1}{3}$
    3. (c)$\frac{1}{2}$
    4. (d)$\frac{3}{4}$
    Show answer

    (c) $\frac{1}{2}$

    Step-by-step solution →
13. Q13
    If $\sin \theta -\cos \theta =0$, ​then the ​value of $({\sin }^4\theta +{\cos }^4\theta )$ ​is

    1. (a)$1$
    2. (b)$\frac{3}{4}$
    3. (c)$\frac{1}{2}$
    4. (d)$\frac{1}{4}$
    Show answer

    (c) $\frac{1}{2}$

    Step-by-step solution →
14. Q14
    $\sin (45^{\circ }+\theta )-\cos (45^{\circ }-\theta )$ is ​equal ​to

    1. (a)$2\cos \theta$
    2. (b)$0$
    3. (c)$2\sin \theta$
    4. (d)$1$
    Show answer

    (b) $0$

    Step-by-step solution →
15. Q15
    A pole 6 ​m high ​casts a ​shadow $2\sqrt{3}$ ​m ​long ​on ​the ground, then the Sun's elevation is

    1. (a)$60^{\circ }$
    2. (b)$45^{\circ }$
    3. (c)$30^{\circ }$
    4. (d)$90^{\circ }$
    Show answer

    (a) $60^{\circ }$

    Step-by-step solution →