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 10 of 23

1. Q181 CBSE 2025
   Prove ​that: $(1+\sin A)(1-\sin A)(1+{\tan }^{2}A)=1$.
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   LHS $=(1-{\sin }^{2}A)(1+{\tan }^{2}A)={\cos }^{2}A\times {\sec }^{2}A={\cos }^{2}A\times \frac{1}{{\cos }^{2}A}=1=$ RHS. Hence proved.

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2. Q182 CBSE 2025
   If $(\sec A+\tan A)(1-\sin A)=k\cos A$, ​then find ​the value ​of $k$.
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   $k=1$

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3. Q183 CBSE 2025
   Evaluate: $\frac{5{\tan }^{2}30^{\circ }+3{\cos }^{2}45^{\circ }-4{\sin }^{2}30^{\circ }}{\sqrt{3}\sin 60^{\circ }\cos 60^{\circ }+{\cot }^{2}45^{\circ }}$
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   The value of the expression is $\frac{26}{21}$.

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4. Q184 CBSE 2025
   Verify ​that $\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$ for $A=30^{\circ }$.
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   At $A=30^{\circ }$: LHS $=\sin 60^{\circ }=\frac{\sqrt{3}}{2}$ and RHS $=\frac{2\tan 30^{\circ }}{1+{\tan }^{2}30^{\circ }}=\frac{2/\sqrt{3}}{4/3}=\frac{\sqrt{3}}{2}$. So LHS $=$ RHS. Hence verified.

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5. Q185 CBSE 2025
   If $3\cot A=4$, ​then determine ​the ​value ​of ${\cos }^{2}A-{\sin }^{2}A$.
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   ${\cos }^{2}A-{\sin }^{2}A=\frac{7}{25}$

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6. Q186 CBSE 2025
   If $\sin (2A+3B)=1$ and $\cos (2A-3B)=\frac{\sqrt{3}}{2}$, $0^{\circ }<2A+3B\le 90^{\circ }$, $A>B$, ​then ​find $A$ and $B$.
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   $A=30^{\circ }$ and $B=10^{\circ }$.

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7. Q187 CBSE 2025
   Prove ​that $\frac{1}{{\cot }^{2}A}+\frac{1}{1+{\tan }^{2}A}=\frac{1}{1-{\sin }^{2}A}-\frac{1}{{\csc }^{2}A}$.
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   Both sides simplify to ${\tan }^{2}A+{\cos }^{2}A$, so LHS = RHS. Hence proved.

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8. Q188 CBSE 2025
   Prove ​that $\frac{\sin \theta -\cos \theta +1}{\cos \theta +\sin \theta -1}=\frac{1}{\sec \theta -\tan \theta }$.
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   LHS $=\sec \theta +\tan \theta =\frac{1}{\sec \theta -\tan \theta }=$ RHS. Hence proved.

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9. Q189 CBSE 2025
   Prove ​that ​: $\left(1+\frac{1}{{\tan }^2\theta }\right)\left(1+\frac{1}{{\cot }^2\theta }\right)=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$.
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   The left side reduces to ${\mathrm{cosec}}^2\theta {\sec }^2\theta =\frac{1}{{\sin }^2\theta {\cos }^2\theta }=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$, which is the right side. Hence proved.

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10. Q190 CBSE 2025
    Prove ​that : $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \times \csc \theta$
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    The left side simplifies to $\frac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }=1+\sec \theta \mathrm{cosec}\theta$, which is the right side. Hence proved.

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11. Q191 CBSE 2025
    If $\sin A+\cos A=\sqrt{3}$, ​then ​prove that $\tan A+\cot A=1$.
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    $\tan A+\cot A=\frac{1}{\sin A\cos A}=\frac{1}{1}=1$. Hence proved.

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12. Q192 CBSE 2025
    Prove ​that : $\sqrt{\frac{\mathrm{cosec}\theta -1}{\mathrm{cosec}\theta +1}}+\sqrt{\frac{\mathrm{cosec}\theta +1}{\mathrm{cosec}\theta -1}}=2\sec \theta$.
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    The left side simplifies to $\frac{2\mathrm{cosec}\theta }{\cot \theta }=2\sec \theta$, which is the right side. Hence proved.

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13. Q193 CBSE 2025
    Prove ​that : $\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2{\sin }^{2}A-1}$
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    The left side simplifies to $\frac{2}{{\sin }^{2}A-{\cos }^{2}A}=\frac{2}{2{\sin }^{2}A-1}$, which is the right side. Hence proved.

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14. Q194 CBSE 2025
    Prove ​that $(\csc A+\sin A{)}^2+(\sec A+\cos A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$.
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    Hence proved: $(\mathrm{cosec}A+\sin A{)}^2+(\sec A+\cos A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$.

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15. Q195 CBSE 2025
    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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16. Q196 CBSE 2025
    Prove that ​: $\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}=2\csc A$.
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    The left side simplifies to $\frac{2\sec A}{\tan A}=2\mathrm{cosec}A$, which is the right side. Hence proved.

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17. Q197 CBSE 2025
    Prove ​that ​: $\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{1}{\sin A}-\sin A\right)=\frac{1}{\tan A+\cot A}$.
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    Both sides simplify to $\sin A\cos A$, so the identity holds. Hence proved.

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18. Q198 CBSE 2025
    Prove ​that: $\frac{1+\csc A}{\csc A}=\frac{{\cos }^{2}A}{1-\sin A}$
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    $\frac{1+\mathrm{cosec}A}{\mathrm{cosec}A}=1+\sin A=\frac{(1+\sin A)(1-\sin A)}{1-\sin A}=\frac{1-{\sin }^{2}A}{1-\sin A}=\frac{{\cos }^{2}A}{1-\sin A}$. Hence proved.

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19. Q199 CBSE 2025
    Prove ​that $\frac{\cos A+\sin A-1}{\cos A-\sin A+1}=\text{cosec }A-\cot A$
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    Dividing the numerator and denominator by $\sin A$ and writing $1={\mathrm{cosec}}^{2}A-{\cot }^{2}A$ in the numerator reduces the left side to $\mathrm{cosec}A-\cot A$; hence the identity is proved.

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20. Q200 CBSE 2025
    If $\cot \theta +\cos \theta =p$ ​and $\cot \theta -\cos \theta =q$, ​prove that $p^2-q^2=4\sqrt{pq}$.
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    Both sides reduce to $4\cot \theta \cos \theta$, since $p^2-q^2=(p+q)(p-q)=2\cot \theta \times 2\cos \theta$ and $pq={\cos }^2\theta {\cot }^2\theta$; hence $p^2-q^2=4\sqrt{pq}$.

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