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 23 of 23

1. Q441 CBSE 2019
   If $\sec \theta +\tan \theta =m$, show ​that $\frac{m^2-1}{m^2+1}=\sin \theta$.
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   Proved: $\frac{m^2-1}{m^2+1}=\frac{2\tan \theta m}{2\sec \theta m}=\frac{\tan \theta }{\sec \theta }=\sin \theta$.

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2. Q442 CBSE 2019
   If $1+{\sin }^2\theta =3\sin \theta \cos \theta$, then ​prove ​that $\tan \theta =1$ ​or $\tan \theta =\frac{1}{2}$.
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   Hence $\tan \theta =1$ or $\tan \theta =\frac{1}{2}$, as required.

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3. Q443 CBSE 2019
   Prove ​that: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$
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   LHS simplifies to $1+\sec \theta \csc \theta =$ RHS, so the identity is proved.

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4. Q444 CBSE 2019
   Prove that ​: $\frac{\sin \theta }{\cot \theta +\mathrm{cosec}\theta }=2+\frac{\sin \theta }{\cot \theta -\mathrm{cosec}\theta }$
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   The difference of the two fractions simplifies to $2$, so $\frac{\sin \theta }{\cot \theta +\mathrm{cosec}\theta }=2+\frac{\sin \theta }{\cot \theta -\mathrm{cosec}\theta }$ is proved.

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5. Q445 CBSE 2019
   Prove ​that ​: $\frac{(1+\cot \theta +\tan \theta )(\sin \theta -\cos \theta )}{({\sec }^3\theta -{\csc }^3\theta )}={\sin }^2\theta {\cos }^2\theta$
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   Identity proved: the left side simplifies to ${\sin }^2\theta {\cos }^2\theta$.

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6. Q446 CBSE 2018
   What is ​the ​value of $({\cos }^{2}67^{\circ }-{\sin }^{2}23^{\circ })$ ?
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   ${\cos }^{2}67^{\circ }-{\sin }^{2}23^{\circ }=0$.

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7. Q447 CBSE 2018
   What ​is ​the ​value of $({\cos }^{2}67^{\circ }-{\sin }^{2}23^{\circ })$?
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   ${\cos }^{2}67^{\circ }-{\sin }^{2}23^{\circ }=0$

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8. Q448 CBSE 2018
   If $\sin \theta +\cos \theta =\sqrt{2}\cos (90^{\circ }-\theta )$, ​find ​the ​value ​of $\cot \theta$.
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   $\cot \theta =\sqrt{2}-1$

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9. Q449 CBSE 2018
   If $4\tan \theta =3$, ​evaluate $\left(\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}\right)$.
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   $\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}=\frac{13}{11}$

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10. Q450 CBSE 2018
    If $\tan \theta +\sin \theta =m$, $\tan \theta -\sin \theta =n$, ​show ​that $m^2-n^2=4\sqrt{mn}$.
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    $m^2-n^2=4\tan \theta \sin \theta =4\sqrt{mn}$. Hence proved.

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11. Q451 CBSE 2018
    If $\tan 2A=\cot (A-18^{\circ })$, ​where $2A$ ​is ​an ​acute ​angle, find the ​value ​of $A$.
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    $A=36^{\circ }$.

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12. Q452 CBSE 2018
    Prove ​that ​: $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$.
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    $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$ (proved by factoring and the identity ${\sin }^{2}A+{\cos }^{2}A=1$).

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13. Q453 CBSE 2018
    Prove that ​: $\left(\frac{\sin A}{1-\cos A}-\frac{1-\cos A}{\sin A}\right)\times \left(\frac{\cos A}{1-\sin A}-\frac{1-\sin A}{\cos A}\right)=4$.
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    $\left(\frac{\sin A}{1-\cos A}-\frac{1-\cos A}{\sin A}\right)\left(\frac{\cos A}{1-\sin A}-\frac{1-\sin A}{\cos A}\right)=2\cot A\times 2\tan A=4$. Hence proved.

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14. Q454 CBSE 2018
    Prove ​that: $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$.
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    $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$, as required.

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