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 17 of 23

1. Q321 CBSE 2023
   Prove ​that ​: $1-\frac{{\sin }^{2}A}{1+\cos A}=\cos A$
   Show answer

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

   Step-by-step solution →
2. Q322 CBSE 2023
   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

   $\frac{67}{12}$

   Step-by-step solution →
3. Q323 CBSE 2023
   For $A=30^{\circ }$ ​and $B=60^{\circ }$, verify ​that $\tan (B-A)=\frac{\tan B-\tan A}{1+\tan A\tan B}$.
   Show answer

   Both sides equal $\frac{1}{\sqrt{3}}$, so the identity is verified.

   Step-by-step solution →
4. Q324 CBSE 2023
   Prove ​that : $\frac{1-{\cot }^2\theta }{1+{\cot }^2\theta }+2{\cos }^2\theta =1$
   Show answer

   $\frac{1-{\cot }^2\theta }{1+{\cot }^2\theta }+2{\cos }^2\theta =1$ (proved).

   Step-by-step solution →
5. Q325 CBSE 2023
   Evaluate: ${\sin }^{2}60^{\circ }-2{\cos }^{2}45^{\circ }+\frac{1}{2}{\mathrm{cosec}}^{2}30^{\circ }$.
   Show answer

   The value is $\frac{7}{4}$.

   Step-by-step solution →
6. Q326 CBSE 2023
   Prove ​that: $\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2\sec A$
   Show answer

   Identity proved: LHS $=2\sec A$.

   Step-by-step solution →
7. Q327 CBSE 2023
   Find ​the ​value of $\frac{3}{4}{\tan }^{2}30^{\circ }-3{\sin }^{2}60^{\circ }+\frac{1}{{\sin }^{2}45^{\circ }}$.
   Show answer

   The value of the expression is $0$.

   Step-by-step solution →
8. Q328 CBSE 2023
   If $a\cos \theta +b\sin \theta =m$ and $a\sin \theta -b\cos \theta =n$, ​then ​prove that $a^2+b^2=m^2+n^2$.
   Show answer

   $a^2+b^2=m^2+n^2$ (proved)

   Step-by-step solution →
9. Q329 CBSE 2023
   Evaluate $\frac{5{\cos }^{2}60^{\circ }+4{\sec }^{2}30^{\circ }-{\tan }^{2}45^{\circ }}{{\sin }^{2}30^{\circ }+{\cos }^{2}30^{\circ }}$.
   Show answer

   $\frac{67}{12}$

   Step-by-step solution →
10. Q330 CBSE 2023
    Prove ​that $\frac{1-\cos \theta }{1+\cos \theta }=(\cot \theta -\csc \theta {)}^2$.
    Show answer

    Proved: $\frac{1-\cos \theta }{1+\cos \theta }=(\cot \theta -\csc \theta {)}^2$

    Step-by-step solution →
11. Q331 CBSE 2023
    If $\tan \theta =\frac{8}{7}$, ​evaluate $\frac{(1-\sin \theta )(1+\sin \theta )}{(1-\cos \theta )(1+\cos \theta )}$.
    Show answer

    $\frac{49}{64}$

    Step-by-step solution →
12. Q332 CBSE 2023
    If $A$ and $B$ ​are acute angles such ​that $\sin (A-B)=0$ ​and $2\cos (A+B)-1=0$, ​then ​find ​angles $A$ ​and $B$.
    Show answer

    $A=30^{\circ },B=30^{\circ }$

    Step-by-step solution →
13. Q333 CBSE 2023
    In ​a right-angled triangle ABC, right ​angled ​at ​C, if $\tan A=\sqrt{3}$, evaluate $\sin A\cos B+\cos A\sin B$.
    Show answer

    1

    Step-by-step solution →
14. Q334 CBSE 2023
    Prove ​that : $\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}=2\mathrm{cosec}A$.
    Show answer

    $2\mathrm{cosec}A$ (proved)

    Step-by-step solution →
15. Q335 CBSE 2023
    Prove ​that $\sec \theta (1-\sin \theta )(\sec \theta +\tan \theta )=1$.
    Show answer

    LHS $=1=$ RHS, proved.

    Step-by-step solution →
16. Q336 CBSE 2023
    Prove ​that $\frac{1+\sec \theta }{\sec \theta }=\frac{{\sin }^2\theta }{1-\cos \theta }$.
    Show answer

    LHS $=1+\cos \theta =$ RHS, proved.

    Step-by-step solution →
17. Q337 CBSE 2023
    Prove ​that $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}={\sec }^{2}A-1$.
    Show answer

    Both sides equal ${\tan }^{2}A$, so the identity is proved.

    Step-by-step solution →
18. Q338 CBSE 2023
    Prove ​that: $\left(\frac{1}{\cos \theta }-\cos \theta \right)\left(\frac{1}{\sin \theta }-\sin \theta \right)=\frac{1}{\tan \theta +\cot \theta }$.
    Show answer

    Both sides reduce to $\sin \theta \cos \theta$, so the identity is proved.

    Step-by-step solution →
19. Q339 CBSE 2023
    Prove ​that $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \mathrm{cosec}\theta$.
    Show answer

    LHS simplifies to $1+\sec \theta \mathrm{cosec}\theta$, proving the identity.

    Step-by-step solution →
20. Q340 CBSE 2023
    Prove ​that ​: $\frac{1-\cos \theta }{1+\cos \theta }=(\csc \theta -\cot \theta {)}^2$
    Show answer

    $\frac{1-\cos \theta }{1+\cos \theta }=(\csc \theta -\cot \theta {)}^2$. Hence proved.

    Step-by-step solution →