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 14 of 23

1. Q261 CBSE 2024
   Evaluate ​: $\sin 30^{\circ }\cos 60^{\circ }+\cos 30^{\circ }\sin 60^{\circ }-\cot 45^{\circ }$
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   $\sin 30^{\circ }\cos 60^{\circ }+\cos 30^{\circ }\sin 60^{\circ }-\cot 45^{\circ }=0$

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2. Q262 CBSE 2024
   Evaluate ​: $\frac{\cos 45^{\circ }+\sin 60^{\circ }}{\sec 30^{\circ }+\mathrm{cosec}30^{\circ }}$.
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   $\frac{\cos 45^{\circ }+\sin 60^{\circ }}{\sec 30^{\circ }+\mathrm{cosec}30^{\circ }}=\frac{3\sqrt{2}+3\sqrt{3}-\sqrt{6}-3}{8}$, which equals $\frac{\sqrt{6}+3}{4(\sqrt{3}+1)}\approx 0.499$.

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3. Q263 CBSE 2024
   If $\cos (A+B)=\frac{1}{2}$ ​and $\tan (A-B)=\frac{1}{\sqrt{3}}$, ​where $0\le A+B\le 90^{\circ }$, ​then ​find the value ​of $\sec (2A-3B)$.
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   $A=45^{\circ }$, $B=15^{\circ }$, so $\sec (2A-3B)=\sec 45^{\circ }=\sqrt{2}$

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4. Q264 CBSE 2024
   Find ​the value ​of $x$ such ​that, $3{\tan }^{2}60^{\circ }-x{\sin }^{2}45^{\circ }+\frac{3}{4}{\sec }^{2}30^{\circ }=2{\mathrm{cosec}}^{2}30^{\circ }$.
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   $x=4$

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5. Q265 CBSE 2024
   Prove ​that $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$.
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   LHS $=\sec \theta +\tan \theta =\frac{1}{\sec \theta -\tan \theta }=$ RHS. Hence proved.

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6. Q266 CBSE 2024
   Prove ​that $\frac{1+\cos \theta +\sin \theta }{1+\cos \theta -\sin \theta }=\frac{1+\sin \theta }{\cos \theta }$.
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   $\text{LHS}=\frac{2(1+\cos \theta )(1+\sin \theta )}{2\cos \theta (1+\cos \theta )}=\frac{1+\sin \theta }{\cos \theta }=\text{RHS}$. Hence proved.

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7. Q267 CBSE 2024
   Prove ​that ​: $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}={\left(\frac{1-\tan A}{1-\cot A}\right)}^2$.
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   Both sides simplify to ${\tan }^{2}A$ (for $0^{\circ }<A<90^{\circ }$, $A\ne 45^{\circ }$). Hence proved.

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8. Q268 CBSE 2024
   Prove ​that : $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \text{cosec}\theta$
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   LHS $=\frac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }=1+\sec \theta \text{cosec}\theta =$ RHS. Hence proved.

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9. Q269 CBSE 2024
   Prove ​that ​: $(\sin \theta +\csc \theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.
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   Hence proved: $(\sin \theta +\mathrm{cosec}\theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.

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10. Q270 CBSE 2024
    If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$, ​then ​show ​that $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$.
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    Hence proved: $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$.

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11. Q271 CBSE 2024
    Prove ​that $\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }=\frac{1-\sin \theta }{\cos \theta }$.
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    LHS $=\sec \theta -\tan \theta =\frac{1-\sin \theta }{\cos \theta }=$ RHS. Hence proved.

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12. Q272 CBSE 2024
    Prove ​that: $\frac{\cot A-\cos A}{\cot A+\cos A}=(\sec A-\tan A{)}^2$.
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    LHS $=\frac{1-\sin A}{1+\sin A}=\frac{(1-\sin A{)}^2}{{\cos }^{2}A}={\left(\frac{1-\sin A}{\cos A}\right)}^2=(\sec A-\tan A{)}^2=$ RHS. Hence proved.

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13. Q273 CBSE 2024
    Prove that ​: $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\sec A\text{cosec}A$
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    LHS $=\frac{1+\sin A\cos A}{\sin A\cos A}=1+\sec A\text{cosec}A=$ RHS. Hence proved.

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14. Q274 CBSE 2024
    Prove ​that $\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$.
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    LHS $=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)}=\frac{\sin \theta (2{\cos }^2\theta -1)}{\cos \theta (2{\cos }^2\theta -1)}=\frac{\sin \theta }{\cos \theta }=\tan \theta =$ RHS. Hence proved.

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15. Q275 CBSE 2024
    Prove ​that $\sqrt{{\sec }^{2}A+{\mathrm{cosec}}^{2}A}=\tan A+\cot A$.
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    Both sides reduce to $\frac{1}{\sin A\cos A}$, so $\sqrt{{\sec }^{2}A+{\mathrm{cosec}}^{2}A}=\tan A+\cot A$. Hence proved.

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16. Q276 CBSE 2024
    Prove ​that : $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$
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    Hence proved: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$ (wherever both sides are defined, i.e. $\theta \ne 45^{\circ }$).

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17. Q277 CBSE 2024
    Prove ​that $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$.
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    LHS $=1+\sec \theta \mathrm{cosec}\theta =$ RHS. Hence proved.

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18. Q278 CBSE 2020
    Prove ​that $(\sin \theta +\csc \theta {)}^2+(\cos \theta +\sec \theta {)}^2=7+{\tan }^2\theta +{\cot }^2\theta$.
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    The left side simplifies to $7+{\tan }^2\theta +{\cot }^2\theta$, equal to the right side; the identity is proved.

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19. Q279 CBSE 2024
    Prove ​that ​: $(\tan A+\sec A{)}^2+(\tan A-\sec A{)}^2=2\left(\frac{1+{\sin }^{2}A}{1-{\sin }^{2}A}\right)$
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    $(\tan A+\sec A{)}^2+(\tan A-\sec A{)}^2=2({\tan }^{2}A+{\sec }^{2}A)=2\left(\frac{1+{\sin }^{2}A}{{\cos }^{2}A}\right)=2\left(\frac{1+{\sin }^{2}A}{1-{\sin }^{2}A}\right)$. Hence proved.

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20. Q280 CBSE 2024
    Prove ​that: $(\cot \theta -\mathrm{cosec}\theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$.
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    $(\cot \theta -\mathrm{cosec}\theta {)}^2=\frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }=\frac{(1-\cos \theta {)}^2}{(1-\cos \theta )(1+\cos \theta )}=\frac{1-\cos \theta }{1+\cos \theta }$. Hence proved.

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