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 21 of 23

1. Q401 CBSE 2020
   Prove ​that ​: $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }={\cos }^2\theta -{\sin }^2\theta$
   Show answer

   $\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }={\cos }^2\theta -{\sin }^2\theta$, hence proved.

   Step-by-step solution →
2. Q402 CBSE 2020
   On the edge $AB$ ​of a wall of ​a house, a projection $AP$, which is perpendicular to the wall, is erected and its edge is of length $27$ cm. A point searchlight is fixed on the corner ​point $P$ of projection ​which sends a ray of ​light $PQ$ ​such that $AQ=9\sqrt{3}$ cm. If $\angle APQ=\theta$, then find ​the value of

   1. (i)$\theta$, and
   2. (ii)$\tan \theta +\sec \theta$.

   Show answer

   (i) $\theta =30^{\circ }$; (ii) $\tan \theta +\sec \theta =\sqrt{3}\approx 1.73$

   Step-by-step solution →
3. Q403 CBSE 2020
   Prove ​that: $(\mathrm{cosec}\theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$
   Show answer

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

   Step-by-step solution →
4. Q404 CBSE 2020
   Prove ​that $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$.
   Show answer

   $\sqrt{\frac{1+\sin A}{1-\sin A}}=\frac{1+\sin A}{\cos A}=\sec A+\tan A$. Hence proved.

   Step-by-step solution →
5. Q405 CBSE 2020
   Prove ​that $\frac{\tan \theta -\cot \theta }{\sin \theta \cos \theta }={\tan }^2\theta -{\cot }^2\theta$.
   Show answer

   RHS $=(\tan \theta -\cot \theta )\frac{1}{\sin \theta \cos \theta }=\frac{\tan \theta -\cot \theta }{\sin \theta \cos \theta }=$ LHS, so the identity is proved.

   Step-by-step solution →
6. Q406 CBSE 2020
   Prove ​that $\frac{\cos \theta }{1-\tan \theta }+\frac{\sin \theta }{1-\cot \theta }=\cos \theta +\sin \theta$
   Show answer

   The left side simplifies to $\cos \theta +\sin \theta$, which equals the right side; hence the identity is proved.

   Step-by-step solution →
7. Q407 CBSE 2020
   Prove ​that $(1+\tan A-\sec A)\times (1+\tan A+\sec A)=2\tan A$.
   Show answer

   Proved: LHS $=(1+\tan A{)}^2-{\sec }^{2}A=2\tan A=$ RHS

   Step-by-step solution →
8. Q408 CBSE 2020
   If $x=3\sin \theta +4\cos \theta$ ​and $y=3\cos \theta -4\sin \theta$ then ​prove that $x^2+y^2=25$.
   Show answer

   $x^2+y^2=25$, hence proved.

   Step-by-step solution →
9. Q409 CBSE 2020
   If $\sin \theta +{\sin }^2\theta =1$; then ​prove ​that ${\cos }^2\theta +{\cos }^4\theta =1$.
   Show answer

   Proved: ${\cos }^2\theta +{\cos }^4\theta =\sin \theta +{\sin }^2\theta =1$.

   Step-by-step solution →
10. Q410 CBSE 2020
    Prove ​that $\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -1}+\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta +1}=2{\sec }^2\theta$.
    Show answer

    LHS simplifies to $2{\sec }^2\theta$, which equals the RHS; hence proved.

    Step-by-step solution →
11. Q411 CBSE 2020
    Prove ​that $\frac{\mathrm{cosec}A}{\mathrm{cosec}A-1}+\frac{\mathrm{cosec}A}{\mathrm{cosec}A+1}=2+2{\tan }^{2}A$.
    Show answer

    LHS simplifies to $2{\sec }^{2}A=2+2{\tan }^{2}A=$ RHS, so the identity is proved.

    Step-by-step solution →
12. Q412 CBSE 2020
    Prove ​that : $$\frac{\sin \theta -\cos \theta +1}{\cos \theta +\sin \theta -1}=\frac{1}{\sec \theta -\tan \theta }$$
    Show answer

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

    Step-by-step solution →
13. Q413 CBSE 2020
    Prove ​that: $(\sin \theta +\mathrm{cosec}\theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.
    Show answer

    Proved: $(\sin \theta +\mathrm{cosec}\theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.

    Step-by-step solution →
14. Q414 CBSE 2020
    Evaluate: $\frac{-\cot (90^{\circ }-\theta )\tan \theta +\sec \theta \mathrm{cosec}(90^{\circ }-\theta )+{\sin }^{2}35^{\circ }+{\sin }^{2}55^{\circ }}{\tan 10^{\circ }\tan 20^{\circ }\tan 30^{\circ }\tan 70^{\circ }\tan 80^{\circ }}$
    Show answer

    $2\sqrt{3}\approx 3.46$

    Step-by-step solution →
15. Q415 CBSE 2020
    If $\sin \theta +\cos \theta =p$ ​and $\sec \theta +\mathrm{cosec}\theta =q$, show ​that $q(p^2-1)=2p$.
    Show answer

    Proved: $q(p^2-1)=2p$.

    Step-by-step solution →
16. Q416 CBSE 2020
    If $\sin \theta +\cos \theta =\sqrt{2}$, prove ​that $\tan \theta +\cot \theta =2$.
    Show answer

    $\tan \theta +\cot \theta =2$ is proved.

    Step-by-step solution →
17. Q417 CBSE 2019
    Find $A$ ​if $\tan 2A=\cot (A-24^{\circ })$.
    Show answer

    $A=38^{\circ }$

    Step-by-step solution →
18. Q418 CBSE 2019
    Find ​the ​value of $({\sin }^{2}33^{\circ }+{\sin }^{2}57^{\circ })$.
    Show answer

    ${\sin }^{2}33^{\circ }+{\sin }^{2}57^{\circ }=1$

    Step-by-step solution →
19. Q419 CBSE 2019
    If $\sin x+\cos y=1$; $x=30^{\circ }$ ​and $y$ ​is ​an acute ​angle, ​find the ​value ​of $y$.
    Show answer

    $y=60^{\circ }$

    Step-by-step solution →
20. Q420 CBSE 2019
    Evaluate ​: ${\sin }^{2}60^{\circ }+2\tan 45^{\circ }-{\cos }^{2}30^{\circ }$
    Show answer

    ${\sin }^{2}60^{\circ }+2\tan 45^{\circ }-{\cos }^{2}30^{\circ }=2$

    Step-by-step solution →