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 11 of 23

1. Q201 CBSE 2025
   Prove ​the ​following trigonometric ​identity ​: $\frac{\cos \theta }{1+\sin \theta }+\frac{1+\sin \theta }{\cos \theta }=2\sec \theta$.
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   $\text{LHS}=\frac{{\cos }^2\theta +(1+\sin \theta {)}^2}{(1+\sin \theta )\cos \theta }=\frac{2(1+\sin \theta )}{(1+\sin \theta )\cos \theta }=\frac{2}{\cos \theta }=2\sec \theta =\text{RHS}$. Hence proved.

   Step-by-step solution →
2. Q202 CBSE 2025
   Prove that ​: $\frac{1+{\cot }^{2}A}{1+{\tan }^{2}A}={\left(\frac{1-\cot A}{1-\tan A}\right)}^2$
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   Both sides reduce to ${\cot }^{2}A$: $\frac{1+{\cot }^{2}A}{1+{\tan }^{2}A}={\cot }^{2}A$ and $\frac{1-\cot A}{1-\tan A}=-\cot A$, whose square is ${\cot }^{2}A$. Hence proved.

   Step-by-step solution →
3. Q203 CBSE 2025
   Prove ​that $2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1=0$.
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   Using ${\sin }^6\theta +{\cos }^6\theta =1-3{\sin }^2\theta {\cos }^2\theta$ and ${\sin }^4\theta +{\cos }^4\theta =1-2{\sin }^2\theta {\cos }^2\theta$, the left-hand side reduces to $0$, which is the right-hand side. Hence proved.

   Step-by-step solution →
4. Q204 CBSE 2025
   Prove ​that: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$.
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   LHS $=1+\tan \theta +\cot \theta =$ RHS. Hence proved.

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5. Q205 CBSE 2025
   Prove ​that $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=\sec \theta \csc \theta +1$.
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   Hence proved: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=\sec \theta \mathrm{cosec}\theta +1$.

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6. Q206 CBSE 2025
   Prove ​that $\frac{\cos \theta -2{\cos }^3\theta }{\sin \theta -2{\sin }^3\theta }+\cot \theta =0$.
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   Taking out $\cos \theta$ and $\sin \theta$ and writing $1={\sin }^2\theta +{\cos }^2\theta$ turns the fraction into $-\cot \theta$, so the left side is $-\cot \theta +\cot \theta =0$. Hence proved.

   Step-by-step solution →
7. Q207 CBSE 2025
   Prove the ​following trigonometric ​identity: $\frac{1+\csc A}{\csc A}=\frac{{\cos }^{2}A}{1-\sin A}$.
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   The left-hand side simplifies to $1+\sin A$, and multiplying and dividing by $1-\sin A$ turns this into $\frac{1-{\sin }^{2}A}{1-\sin A}=\frac{{\cos }^{2}A}{1-\sin A}$, the right-hand side; hence the identity is proved.

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8. Q208 CBSE 2025
   Given that $\sin \theta +\cos \theta =x$, ​prove ​that ${\sin }^4\theta +{\cos }^4\theta =\frac{2-(x^2-1{)}^2}{2}$.
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   Squaring the given relation gives $2\sin \theta \cos \theta =x^2-1$, and substituting this reduces the right side to $1-2{\sin }^2\theta {\cos }^2\theta ={\sin }^4\theta +{\cos }^4\theta$. Hence proved.

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9. Q209 CBSE 2025
   Let $2A+B$ and $A+2B$ ​be acute ​angles ​such that $\sin (2A+B)=\frac{\sqrt{3}}{2}$ and $\tan (A+2B)=1$. ​Find the ​value ​of $\cot (4A-7B)$.
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   Here $A=25^{\circ }$ and $B=10^{\circ }$, so $4A-7B=30^{\circ }$ and $\cot (4A-7B)=\cot 30^{\circ }=\sqrt{3}$.

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10. Q210 CBSE 2025
    Evaluate: $\frac{2}{3}({\cos }^{4}30^{\circ }-{\sin }^{4}45^{\circ })-3({\sin }^{2}60^{\circ }-{\sec }^{2}45^{\circ })+\frac{1}{4}{\cot }^{2}30^{\circ }$.
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    The value of the expression is $\frac{113}{24}$, that is $4\frac{17}{24}$.

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11. Q211 CBSE 2025
    Prove ​that $\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{1-2{\cos }^{2}A}.$
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    $\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{{\sin }^{2}A-{\cos }^{2}A}=\frac{2}{1-2{\cos }^{2}A}$. Hence proved.

    Step-by-step solution →
12. Q212 CBSE 2025
    A teacher ​asked his students ​to draw ​a right triangle $ABC$ ​with $AB=8$ cm, $\angle B=90^{\circ }$ and $BC=15$ cm. ​Based ​on ​the above, answer the following:

    1. (i)Evaluate $({\sin }^{2}A-{\cos }^{2}A)$.[1]
    2. (ii)Evaluate $\left(\frac{1}{{\cos }^{2}A}-\frac{1}{{\cot }^{2}A}\right)$.[1]
    3. (iii)(a)Evaluate $\frac{2\tan A}{1+{\tan }^{2}A}$ and prove that it is equal to $2\sin A\cos A$.[2]
    4. (iii)(b) OREvaluate $\frac{{\tan }^{2}A-{\sec }^{2}A}{{\cot }^{2}A-{\csc }^{2}A}$.[2]
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    (i) $\frac{161}{289}$; (ii) $1$; (iii)(a) $\frac{240}{289}$, which equals $2\sin A\cos A=2\times \frac{15}{17}\times \frac{8}{17}$; hence proved; (iii)(b) $1$

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13. Q213 CBSE 2024
    If $\frac{x}{3}=2\sin A$, $\frac{y}{3}=2\cos A$, then ​the value ​of $x^2+y^2$ ​is :

    1. (a)$36$
    2. (b)$9$
    3. (c)$6$
    4. (d)$18$
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    (a) $36$

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14. Q214 CBSE 2024
    If $4\sec \theta -5=0$, ​then ​the value ​of $\cot \theta$ ​is ​:

    1. (a)$\frac{3}{4}$
    2. (b)$\frac{4}{5}$
    3. (c)$\frac{5}{3}$
    4. (d)$\frac{4}{3}$
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    (d) $\frac{4}{3}$

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15. Q215 CBSE 2024
    If $\sin \theta =\frac{1}{3}$, ​then $\sec \theta$ ​is ​equal ​to:

    1. (a)$\frac{2\sqrt{2}}{3}$
    2. (b)$\frac{3}{2\sqrt{2}}$
    3. (c)$3$
    4. (d)$\frac{1}{\sqrt{3}}$
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    (b) $\frac{3}{2\sqrt{2}}$

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16. Q216 CBSE 2024
    If $\sin A=\frac{3}{5}$, ​then ​the value of $\cot A$ ​is ​:

    1. (a)$\frac{3}{4}$
    2. (b)$\frac{4}{3}$
    3. (c)$\frac{4}{5}$
    4. (d)$\frac{5}{4}$
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    (b) $\frac{4}{3}$

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17. Q217 CBSE 2024
    If $\sec \theta -\tan \theta =m$, ​then ​the value of $\sec \theta +\tan \theta$ ​is ​:

    1. (a)$1-\frac{1}{m}$
    2. (b)$m^2-1$
    3. (c)$\frac{1}{m}$
    4. (d)$-m$
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    (c) $\frac{1}{m}$

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18. Q218 CBSE 2024
    If $\tan A=\frac{3}{4}$, ​then $\frac{{\sin }^{2}A+{\cos }^{2}A}{\sec A}$ is ​equal ​to :

    1. (a)$\frac{4}{3}$
    2. (b)$\frac{4}{5}$
    3. (c)$\frac{3}{5}$
    4. (d)$\frac{5}{4}$
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    (b) $\frac{4}{5}$

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19. Q219 CBSE 2024
    If $\sin \theta =\cos \theta$, $(0^{\circ }<\theta <90^{\circ })$, ​then ​the value ​of $(\sec \theta \times \sin \theta )$ is ​:

    1. (a)$\frac{1}{\sqrt{2}}$
    2. (b)$\sqrt{2}$
    3. (c)$1$
    4. (d)$0$
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    (c) $1$

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20. Q220 CBSE 2024
    $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}$ is ​equal ​to

    1. (a)${\sec }^{2}A$
    2. (b)$-1$
    3. (c)${\cot }^{2}A$
    4. (d)${\tan }^{2}A$
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    (d) ${\tan }^{2}A$

    Step-by-step solution →