Class 10 Maths · Chapter 8

Introduction to Trigonometry

454 Class 10 Maths questions on Introduction to Trigonometry from the NCERT textbook, NCERT Exemplar and CBSE board papers 2018–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 13 of 23

1. Q241 CBSE 2024
   Choose ​the ​correct ​option for the following ​Assertion ​(A) ​and Reason (R).

   Assertion (A): If $\sin A=\frac{1}{3}$ $(0^{\circ }<A<90^{\circ })$, then the value of $\cos A$ is $\frac{2\sqrt{2}}{3}$.

   Reason (R): For every angle $\theta$, ${\sin }^2\theta +{\cos }^2\theta =1$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true, but Reason (R) is false.
   4. (d)Assertion (A) is false, but Reason (R) is true.
   Show answer

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   Step-by-step solution →
2. Q242 CBSE 2024
   Evaluate ​: ${\sin }^{2}30^{\circ }+{\cos }^{2}45^{\circ }-\cos 0^{\circ }\times \tan 45^{\circ }$
   Show answer

   ${\sin }^{2}30^{\circ }+{\cos }^{2}45^{\circ }-\cos 0^{\circ }\times \tan 45^{\circ }=-\frac{1}{4}$

   Step-by-step solution →
3. Q243 CBSE 2024
   If $\cos \theta =\frac{3}{5}$, ​find the ​value of $\frac{\tan \theta -\sin \theta }{\mathrm{cosec}\theta +\cot \theta }$.
   Show answer

   $\frac{\tan \theta -\sin \theta }{\mathrm{cosec}\theta +\cot \theta }=\frac{4}{15}$

   Step-by-step solution →
4. Q244 CBSE 2024
   For $\theta =30^{\circ }$, ​verify ​that $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$.
   Show answer

   At $\theta =30^{\circ }$: LHS $=\cos 60^{\circ }=\frac{1}{2}$ and RHS $=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$. Both sides are equal, so $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$ is verified.

   Step-by-step solution →
5. Q245 CBSE 2024
   Evaluate ​: $4{\sin }^{2}60^{\circ }{\tan }^{2}45^{\circ }-2{\sec }^{2}30^{\circ }{\tan }^{2}60^{\circ }$
   Show answer

   $4{\sin }^{2}60^{\circ }{\tan }^{2}45^{\circ }-2{\sec }^{2}30^{\circ }{\tan }^{2}60^{\circ }=-5$

   Step-by-step solution →
6. Q246 CBSE 2024
   Evaluate ​: $\frac{2\tan 30^{\circ }\times \sec 60^{\circ }\times \tan 45^{\circ }}{1-{\sin }^{2}60^{\circ }}$
   Show answer

   $\frac{16\sqrt{3}}{3}$ (equal to $\frac{16}{\sqrt{3}}\approx 9.24$)

   Step-by-step solution →
7. Q247 CBSE 2024
   If $12\mathrm{cosec}A=13$, ​then ​find ​the ​value of $\frac{2\sin A-3\cos A}{4\sin A-9\cos A}$.
   Show answer

   The value of the expression is $3$.

   Step-by-step solution →
8. Q248 CBSE 2024
   If $\mathrm{cosec}\theta =\sqrt{10}$, ​then ​find the ​values ​of $\tan \theta$ and $\sec \theta$.
   Show answer

   $\tan \theta =\frac{1}{3}$ and $\sec \theta =\frac{\sqrt{10}}{3}$

   Step-by-step solution →
9. Q249 CBSE 2024
   Evaluate ​: $\frac{5{\cos }^{2}60^{\circ }+4{\sec }^{2}30^{\circ }-{\tan }^{2}45^{\circ }}{{\sin }^{2}30^{\circ }+{\sin }^{2}60^{\circ }}$
   Show answer

   The value is $\frac{67}{12}$ $\left(=5\frac{7}{12}\right)$.

   Step-by-step solution →
10. Q250 CBSE 2024
    Evaluate ​: $2{\sin }^{2}30^{\circ }\sec 60^{\circ }+{\tan }^{2}60^{\circ }$.
    Show answer

    $2{\sin }^{2}30^{\circ }\sec 60^{\circ }+{\tan }^{2}60^{\circ }=4$

    Step-by-step solution →
11. Q251 CBSE 2024
    Evaluate ​: $2\sqrt{2}\cos 45^{\circ }\sin 30^{\circ }+2\sqrt{3}\cos 30^{\circ }$
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    $2\sqrt{2}\cos 45^{\circ }\sin 30^{\circ }+2\sqrt{3}\cos 30^{\circ }=4$

    Step-by-step solution →
12. Q252 CBSE 2024
    If $A=60^{\circ }$ and $B=30^{\circ }$, verify ​that ​: $\sin (A+B)=\sin A\cos B+\cos A\sin B$
    Show answer

    LHS $=\sin 90^{\circ }=1$ and RHS $=\frac{3}{4}+\frac{1}{4}=1$, so LHS $=$ RHS. Hence verified.

    Step-by-step solution →
13. Q253 CBSE 2024
    If $2\sin (A+B)=\sqrt{3}$ ​and $\cos (A-B)=1$, ​then ​find ​the ​measures ​of angles $A$ and $B$. $0^{\circ }\le A,B,(A+B)\le 90^{\circ }$.
    Show answer

    $A=30^{\circ }$ and $B=30^{\circ }$.

    Step-by-step solution →
14. Q254 CBSE 2024
    If $\sin (A-B)=\frac{1}{2}$, $\cos (A+B)=\frac{1}{2}$; $0<A+B\le 90^{\circ }$, $A>B$; ​find $\angle A$ ​and $\angle B$.
    Show answer

    $\angle A=45^{\circ }$ and $\angle B=15^{\circ }$.

    Step-by-step solution →
15. Q255 CBSE 2024
    Prove ​that ${\cos }^{4}A-{\sin }^{4}A=2{\cos }^{2}A-1$.
    Show answer

    ${\cos }^{4}A-{\sin }^{4}A=({\cos }^{2}A-{\sin }^{2}A)({\cos }^{2}A+{\sin }^{2}A)={\cos }^{2}A-{\sin }^{2}A=2{\cos }^{2}A-1$. Hence proved.

    Step-by-step solution →
16. Q256 CBSE 2024
    Evaluate: $\sin A\cos B+\cos A\sin B$; ​if $A=30^{\circ }$ ​and $B=45^{\circ }$.
    Show answer

    $\sin A\cos B+\cos A\sin B=\frac{1+\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}+\sqrt{6}}{4}\approx 0.966$

    Step-by-step solution →
17. Q257 CBSE 2024
    If $\tan A=\sqrt{3}$, ​then find ​the value ​of ${\cos }^{2}A-{\sin }^{2}A$.
    Show answer

    ${\cos }^{2}A-{\sin }^{2}A=-\frac{1}{2}$

    Step-by-step solution →
18. Q258 CBSE 2024
    If $\tan \theta +\sec \theta =m$, ​then ​prove that $\sec \theta =\frac{m^2+1}{2m}$.
    Show answer

    $\sec \theta =\frac{m^2+1}{2m}$. Hence proved.

    Step-by-step solution →
19. Q259 CBSE 2024
    If $\sin A=\frac{3}{5}$ ​and $\cos B=\frac{12}{13}$, then ​find the ​value of $(\tan A+\tan B)$.
    Show answer

    $\tan A+\tan B=\frac{7}{6}$

    Step-by-step solution →
20. Q260 CBSE 2024
    If $x\sin 60^{\circ }+\cos 30^{\circ }-\tan 45^{\circ }=\frac{\sqrt{3}}{2}$, ​find the value ​of $x$.
    Show answer

    $x=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$

    Step-by-step solution →