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 20 of 23

1. Q381 CBSE 2020
   Evaluate ​: $(\tan 23^{\circ })\times (\tan 67^{\circ })$
   Show answer

   $\tan 23^{\circ }\times \tan 67^{\circ }=1$.

   Step-by-step solution →
2. Q382 CBSE 2020
   If $\mathrm{cosec}\theta =\frac{5}{4}$, find ​the ​value ​of $\cot \theta$.
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   $\cot \theta =\frac{3}{4}$.

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3. Q383 CBSE 2020
   Find ​the ​value of $\sin 42^{\circ }-\cos 48^{\circ }$.
   Show answer

   $\sin 42^{\circ }-\cos 48^{\circ }=0$.

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4. Q384 CBSE 2020
   Find the ​value ​of $\cos 0^{\circ }\times \cos 30^{\circ }\times \cos 45^{\circ }\times \cos 60^{\circ }$.
   Show answer

   $\frac{\sqrt{6}}{8}$.

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5. Q385 CBSE 2020
   Evaluate ​: $\frac{2\tan 45^{\circ }\times \cos 60^{\circ }}{\sin 30^{\circ }}$
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   $\frac{2\tan 45^{\circ }\times \cos 60^{\circ }}{\sin 30^{\circ }}=2$

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6. Q386 CBSE 2020
   Find the ​value of $\sin 30^{\circ }\times \cos 45^{\circ }+\sin 45^{\circ }\times \cos 30^{\circ }$.
   Show answer

   $\sin 30^{\circ }\cos 45^{\circ }+\sin 45^{\circ }\cos 30^{\circ }=\frac{\sqrt{2}+\sqrt{6}}{4}(=\sin 75^{\circ }\approx 0.966)$.

   Step-by-step solution →
7. Q387 CBSE 2020
   Prove ​that $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$.
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   Proved: $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\frac{1-\sin \theta }{\cos \theta }=\sec \theta -\tan \theta$

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8. Q388 CBSE 2020
   Prove ​that $\frac{{\tan }^2\theta }{1+{\tan }^2\theta }+\frac{{\cot }^2\theta }{1+{\cot }^2\theta }=1$.
   Show answer

   Proved: LHS $={\sin }^2\theta +{\cos }^2\theta =1=$ RHS

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9. Q389 CBSE 2020
   If $\sec 2A=\mathrm{cosec}(A-30^{\circ })$, $0^{\circ }<2A<90^{\circ }$, ​then ​find ​the ​value of $\angle A$.
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   $\angle A=40^{\circ }$.

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10. Q390 CBSE 2020
    If $A$, $B$ ​and $C$ ​are ​interior ​angles ​of ​a $△ABC$, then show that $\cos \left(\frac{B+C}{2}\right)=\sin \left(\frac{A}{2}\right)$.
    Show answer

    Since $\frac{B+C}{2}=90^{\circ }-\frac{A}{2}$, we have $\cos \left(\frac{B+C}{2}\right)=\sin \left(\frac{A}{2}\right)$, as required.

    Step-by-step solution →
11. Q391 CBSE 2020
    Evaluate ​: ${\left(\frac{\sin 47^{\circ }}{\cos 43^{\circ }}\right)}^2+{\left(\frac{\cos 30^{\circ }}{\cot 30^{\circ }}\right)}^2-(\sin 60^{\circ }{)}^2$
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    The value of the expression is $\frac{1}{2}$.

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12. Q392 CBSE 2020
    The rod $AC$ of ​a TV disc antenna is ​fixed at right angles to the wall $AB$ and a ​rod $CD$ ​is supporting the ​disc as shown in Fig. 4. If $AC=1.5$ ​m long and $CD=3$ m, ​find ​

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

    Show answer

    (i) $\tan \theta =\frac{1}{\sqrt{3}}$; (ii) $\sec \theta +\csc \theta =2+\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}+6}{3}\approx 3.15$

    Step-by-step solution →
13. Q393 CBSE 2020
    If $\tan \theta =\frac{3}{4}$, ​find ​the value ​of $\left(\frac{1-{\cos }^2\theta }{1+{\cos }^2\theta }\right)$.
    Show answer

    $\frac{1-{\cos }^2\theta }{1+{\cos }^2\theta }=\frac{9}{41}$

    Step-by-step solution →
14. Q394 CBSE 2020
    Prove ​that $1+\frac{{\cot }^2\alpha }{1+\mathrm{cosec}\alpha }=\mathrm{cosec}\alpha$.
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    LHS simplifies to $\mathrm{cosec}\alpha =$ RHS, so the identity is proved.

    Step-by-step solution →
15. Q395 CBSE 2020
    Show ​that $\sec 41^{\circ }\times \sin 49^{\circ }+\cos 49^{\circ }\times \mathrm{cosec}41^{\circ }=2$.
    Show answer

    LHS $=1+1=2=$ RHS. Hence proved.

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16. Q396 CBSE 2020
    If $\tan \theta =\sqrt{3}$, ​find ​the value ​of $\left(\frac{2\sec \theta }{1+{\tan }^2\theta }\right)$.
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    $1$

    Step-by-step solution →
17. Q397 CBSE 2020
    If $\cos (A-B)=\sin (A+B)=1$, then ​find the ​values ​of $A$ ​and $B$.
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    $A=45^{\circ }$ and $B=45^{\circ }$.

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18. Q398 CBSE 2020
    Prove that ​: $\sqrt{{\sec }^2\theta +{\csc }^2\theta }=\tan \theta +\cot \theta$
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    Both sides equal $\frac{1}{\sin \theta \cos \theta }$, hence $\sqrt{{\sec }^2\theta +{\csc }^2\theta }=\tan \theta +\cot \theta$ is proved.

    Step-by-step solution →
19. Q399 CBSE 2020
    Prove ​that ​: $\frac{1}{1+\sin \theta }+\frac{1}{1-\sin \theta }=2{\sec }^2\theta$
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    $\frac{1}{1+\sin \theta }+\frac{1}{1-\sin \theta }=2{\sec }^2\theta$, hence proved.

    Step-by-step solution →
20. Q400 CBSE 2020
    Prove ​that: $\frac{\sin \theta }{1-\cos \theta }=(\mathrm{cosec}\theta +\cot \theta )$.
    Show answer

    Starting from the LHS and rationalising gives $\mathrm{cosec}\theta +\cot \theta$, which equals the RHS; hence the identity is proved.

    Step-by-step solution →