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 22 of 23

1. Q421 CBSE 2019
   If $\sin A=\frac{3}{4}$, ​calculate $\sec A$.
   Show answer

   $\sec A=\frac{4}{\sqrt{7}}=\frac{4\sqrt{7}}{7}\approx 1.51$

   Step-by-step solution →
2. Q422 CBSE 2019
   If $\cot A+\frac{1}{\cot A}=2$, ​then find ​the value ​of ${\cot }^{2}A+\frac{1}{{\cot }^{2}A}$.
   Show answer

   ${\cot }^{2}A+\frac{1}{{\cot }^{2}A}=2$.

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3. Q423 CBSE 2019
   Evaluate: $\frac{\tan 65^{\circ }}{\cot 25^{\circ }}$.
   Show answer

   $\frac{\tan 65^{\circ }}{\cot 25^{\circ }}=1$.

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4. Q424 CBSE 2019
   If $\sin (20^{\circ }+\theta )=\cos 30^{\circ }$, then ​find ​the ​value of $\theta$.
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   $\theta =40^{\circ }$.

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5. Q425 CBSE 2019
   Express $(\sin 67^{\circ }+\cos 75^{\circ })$ in ​terms of ​trigonometric ​ratios of ​the ​angle ​between $0^{\circ }$ and $45^{\circ }$.
   Show answer

   $\sin 67^{\circ }+\cos 75^{\circ }=\cos 23^{\circ }+\sin 15^{\circ }$.

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6. Q426 CBSE 2019
   In ​Figure 1, $PS=3$ ​cm, $QS=4$ cm, $\angle PRQ=\theta$, $\angle PSQ=90^{\circ }$, $PQ\perp RQ$ ​and $RQ=9$ cm. ​Evaluate $\tan \theta$.

   Show answer

   $\tan \theta =\frac{5}{9}$

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7. Q427 CBSE 2019
   If $\tan \alpha =\frac{5}{12}$, ​find ​the ​value of $\sec \alpha$.
   Show answer

   $\sec \alpha =\frac{13}{12}$.

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8. Q428 CBSE 2019
   Prove ​that: $(\sin \theta +1+\cos \theta )(\sin \theta -1+\cos \theta )\times \sec \theta \csc \theta =2$.
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   The left-hand side simplifies to $2$, so the identity is proved.

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9. Q429 CBSE 2019
   Prove ​that $\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}=2\mathrm{cosec}\theta$.
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   Proved: the left-hand side simplifies to $\frac{2\sec \theta }{\tan \theta }=2\mathrm{cosec}\theta$.

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10. Q430 CBSE 2019
    Prove ​that ​: $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\mathrm{cosec}\theta$
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    Proved: $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\mathrm{cosec}\theta$.

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11. Q431 CBSE 2019
    Prove ​that: $\frac{1}{\csc \theta +\cot \theta }-\frac{1}{\sin \theta }=\frac{1}{\sin \theta }-\frac{1}{\csc \theta -\cot \theta }$
    Show answer

    Both sides simplify to $-\cot \theta$, hence $\text{LHS}=\text{RHS}$. Proved.

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12. Q432 CBSE 2019
    Prove ​that ​: $\frac{\tan \theta }{1-\tan \theta }-\frac{\cot \theta }{1-\cot \theta }=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$
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    Identity proved: $\frac{\tan \theta }{1-\tan \theta }-\frac{\cot \theta }{1-\cot \theta }=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$.

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13. Q433 CBSE 2019
    If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$, ​show ​that $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$.
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    $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$. Hence proved.

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14. Q434 CBSE 2019
    Evaluate: ${\left(\frac{3\sin 43^{\circ }}{\cos 47^{\circ }}\right)}^2-\frac{\cos 37^{\circ }\csc 53^{\circ }}{\tan 5^{\circ }\tan 25^{\circ }\tan 45^{\circ }\tan 65^{\circ }\tan 85^{\circ }}$
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    $8$

    Step-by-step solution →
15. Q435 CBSE 2019
    Prove ​that $(\sin \theta +\mathrm{cosec}\theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.
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    Proved: $(\sin \theta +\mathrm{cosec}\theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.

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16. Q436 CBSE 2019
    Prove ​that $(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)=2$.
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    $(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)=2$. Proved.

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17. Q437 CBSE 2019
    $A$, $B$ ​and $C$ are interior angles of ​a ​triangle $ABC$. Show that

    1. (i)$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
    2. (ii)​If $\angle A=90^{\circ }$, ​then ​find the ​value of $\tan \left(\frac{B+C}{2}\right)$.
    Show answer

    (i) $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$ is proved. (ii) $\tan \left(\frac{B+C}{2}\right)=1$.

    Step-by-step solution →
18. Q438 CBSE 2019
    If $\tan (A+B)=1$ and $\tan (A-B)=\frac{1}{\sqrt{3}}$, $0^{\circ }<A+B<90^{\circ }$, $A>B$, ​then find ​the ​values ​of $A$ ​and $B$.
    Show answer

    $A=37.5^{\circ }$ and $B=7.5^{\circ }$.

    Step-by-step solution →
19. Q439 CBSE 2019
    Prove ​that ​: $\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }=\sec \theta \mathrm{cosec}\theta -2\sin \theta \cos \theta$
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    Proved: $\frac{{\tan }^3\theta }{1+{\tan }^2\theta }+\frac{{\cot }^3\theta }{1+{\cot }^2\theta }=\sec \theta \mathrm{cosec}\theta -2\sin \theta \cos \theta$.

    Step-by-step solution →
20. Q440 CBSE 2019
    Prove ​that $\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}$.
    Show answer

    $\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\sec A+\tan A=\frac{1}{\sec A-\tan A}$. Proved.

    Step-by-step solution →