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 6 of 23

1. Q101 CBSE 2026
   Prove ​that ​: $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$
   Show answer

   Proved: $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$.

   Step-by-step solution →
2. Q102 CBSE 2026
   If $\sin (A+2B)=\frac{\sqrt{3}}{2}$ ​and $\cos (A+4B)=0$, $A>B$, ​find $A$ ​and $B$.
   Show answer

   $A=30^{\circ }$ and $B=15^{\circ }$ (which satisfies $A>B$).

   Step-by-step solution →
3. Q103 CBSE 2026
   For acute angles $A$ and $B$, and $A+2B$ and $2A+B$ are ​acute, ​if $\tan (A+2B)=\sqrt{3}$ and $\sin (2A+B)=\frac{1}{\sqrt{2}}$, then ​find the ​measures of angles $A$ ​and $B$.
   Show answer

   $A=10^{\circ }$ and $B=25^{\circ }$.

   Step-by-step solution →
4. Q104 CBSE 2026
   If $4\tan A=3$, then ​find ​the ​value ​of $\frac{{\mathrm{cosec}}^{2}A+1}{{\mathrm{cosec}}^{2}A-1}$.
   Show answer

   $\frac{{\mathrm{cosec}}^{2}A+1}{{\mathrm{cosec}}^{2}A-1}=\frac{17}{8}$.

   Step-by-step solution →
5. Q105 CBSE 2026
   If $\sin (A+2B)=2\cos 60^{\circ }$ and $A=3B$, ​find ​the ​measures of $A$ ​and $B$.
   Show answer

   $A=54^{\circ }$ and $B=18^{\circ }$.

   Step-by-step solution →
6. Q106 CBSE 2026
   Evaluate ​: $\frac{2{\sin }^{2}60^{\circ }+{\cos }^{2}60^{\circ }}{{\tan }^{2}30^{\circ }}$
   Show answer

   $\frac{2{\sin }^{2}60^{\circ }+{\cos }^{2}60^{\circ }}{{\tan }^{2}30^{\circ }}=\frac{21}{4}$

   Step-by-step solution →
7. Q107 CBSE 2026
   It ​is ​given that $\sin (A-B)=\sin A\cos B-\cos A\sin B$. ​Use ​it ​to ​evaluate $\sin 15^{\circ }$.
   Show answer

   $\sin 15^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

   Step-by-step solution →
8. Q108 CBSE 2026
   Prove that ​: $(\sin A+\sec A{)}^2+(\cos A+\csc A{)}^2=(1+\sec A\csc A{)}^2$
   Show answer

   Proved. The left-hand side simplifies to $1+2\sec A\csc A+{\sec }^{2}A{\csc }^{2}A$, which is exactly $(1+\sec A\csc A{)}^2$, the right-hand side.

   Step-by-step solution →
9. Q109 CBSE 2026
   Prove ​that ​: $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$
   Show answer

   Both sides reduce to $\sec \theta +\tan \theta$, so $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$ is proved.

   Step-by-step solution →
10. Q110 CBSE 2026
    If $x=h+a\cos \theta$, $y=k+b\sin \theta$, ​then ​prove ​that ​: ${\left(\frac{x-h}{a}\right)}^2+{\left(\frac{y-k}{b}\right)}^2=1$
    Show answer

    By writing $\cos \theta =\frac{x-h}{a}$ and $\sin \theta =\frac{y-k}{b}$, squaring and adding, the identity ${\sin }^2\theta +{\cos }^2\theta =1$ gives ${\left(\frac{x-h}{a}\right)}^2+{\left(\frac{y-k}{b}\right)}^2=1$. Hence proved.

    Step-by-step solution →
11. Q111 CBSE 2026
    Prove ​that ​: $\frac{{\sec }^3\theta }{{\sec }^2\theta -1}+\frac{{\mathrm{cosec}}^3\theta }{{\mathrm{cosec}}^2\theta -1}=\sec \theta \cdot \mathrm{cosec}\theta (\sec \theta +\mathrm{cosec}\theta )$
    Show answer

    It is proved that $\frac{{\sec }^3\theta }{{\sec }^2\theta -1}+\frac{{\mathrm{cosec}}^3\theta }{{\mathrm{cosec}}^2\theta -1}=\sec \theta \cdot \mathrm{cosec}\theta (\sec \theta +\mathrm{cosec}\theta )$.

    Step-by-step solution →
12. Q112 CBSE 2026
    Prove ​that $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\sec A-\tan A}{\sec A+\tan A}$.
    Show answer

    Both sides simplify to $\frac{1-\sin A}{1+\sin A}$, so $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\sec A-\tan A}{\sec A+\tan A}$. Hence proved.

    Step-by-step solution →
13. Q113 CBSE 2024
    Prove ​that $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$.
    Show answer

    Hence proved: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$ (for $\theta \ne 45^{\circ }$).

    Step-by-step solution →
14. Q114 CBSE 2026
    Prove that ​: $\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2\mathrm{cosec}A$
    Show answer

    It is proved that $\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2\mathrm{cosec}A$, by combining the fractions and using $1-{\sec }^{2}A=-{\tan }^{2}A$.

    Step-by-step solution →
15. Q115 CBSE 2026
    If $\frac{\sec \alpha }{\mathrm{cosec}\beta }=p$ ​and $\frac{\tan \alpha }{\mathrm{cosec}\beta }=q$, then ​prove that $(p^2-q^2){\sec }^2\alpha =p^2$.
    Show answer

    It is proved that $(p^2-q^2){\sec }^2\alpha =p^2$, since both sides equal ${\sec }^2\alpha {\sin }^2\beta$.

    Step-by-step solution →
16. Q116 CBSE 2026
    Prove ​that $(\mathrm{cosec}A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$.
    Show answer

    Both sides simplify to $\sin A\cos A$, so $(\mathrm{cosec}A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$ is proved.

    Step-by-step solution →
17. Q117 CBSE 2026
    Prove ​that $\sqrt{\frac{1-\sin A}{1+\sin A}}=\frac{1}{\sec A+\tan A}$.
    Show answer

    Both sides equal $\sec A-\tan A$, so $\sqrt{\frac{1-\sin A}{1+\sin A}}=\frac{1}{\sec A+\tan A}$. Hence proved.

    Step-by-step solution →
18. Q118 CBSE 2026
    Prove ​that : $\frac{\sin A-\tan A}{\sin A+\tan A}=\frac{1-\sec A}{1+\sec A}$
    Show answer

    $\frac{\sin A-\tan A}{\sin A+\tan A}=\frac{1-\sec A}{1+\sec A}$ is proved by factoring $\sin A$ and using $\sec A=\frac{1}{\cos A}$

    Step-by-step solution →
19. Q119 CBSE 2026
    If $\sin A+{\sin }^{2}A=1$, ​find ​the ​value ​of ${\cos }^{2}A+{\cos }^{4}A$. Also, ​using ​the above, prove that ${\tan }^{2}A\times {\sec }^{2}A=1$.
    Show answer

    ${\cos }^{2}A+{\cos }^{4}A=1$, and hence ${\tan }^{2}A\times {\sec }^{2}A=\frac{{\sin }^{2}A}{{\cos }^{4}A}=1$ - proved

    Step-by-step solution →
20. Q120 CBSE 2026
    Prove ​that : $(\sin A-\text{cosec}A{)}^2+(\cos A-\sec A{)}^2={\tan }^{2}A+{\cot }^{2}A-1$
    Show answer

    LHS $={\tan }^{2}A+{\cot }^{2}A-1=$ RHS, so the identity is proved.

    Step-by-step solution →