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 9 of 23

1. Q161 CBSE 2025
   Read the ​assertion ​and ​reason and choose ​the ​correct ​option.

   Assertion (A): In a right angle triangle $ABC$, $\angle B=90^{\circ }$. Therefore the value of $\cos (A+C)$ is equal to $0$.

   Reason (R): $A+B+C=180^{\circ }$ and $\cos 90^{\circ }=0$.

   1. (a)Both A and R are true and R is the correct explanation of A.
   2. (b)Both A and R are true but R is not the correct explanation of A.
   3. (c)A is true but R is false.
   4. (d)A is false but R is true.
   Show answer

   (a) Both A and R are true and R is the correct explanation of A.

   Step-by-step solution →
2. Q162 CBSE 2025
   If $4k={\tan }^{2}60^{\circ }-2{\csc }^{2}30^{\circ }-2{\tan }^{2}30^{\circ }$, ​then find the ​value ​of $k$.
   Show answer

   The value of $k$ is $-\frac{17}{12}$.

   Step-by-step solution →
3. Q163 CBSE 2025
   If $x\cos 60^{\circ }+y\cos 0^{\circ }+\sin 30^{\circ }-\cot 45^{\circ }=5$, then ​find ​the ​value ​of $x+2y$.
   Show answer

   The value of $x+2y$ is $11$.

   Step-by-step solution →
4. Q164 CBSE 2025
   Evaluate: $\frac{{\tan }^{2}60^{\circ }}{{\sin }^{2}60^{\circ }+{\cos }^{2}30^{\circ }}$
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   The value of $\frac{{\tan }^{2}60^{\circ }}{{\sin }^{2}60^{\circ }+{\cos }^{2}30^{\circ }}$ is $2$.

   Step-by-step solution →
5. Q165 CBSE 2025
   Evaluate: $2{\tan }^{2}45^{\circ }+{\cos }^{2}30^{\circ }-{\sin }^{2}90^{\circ }$.
   Show answer

   $2{\tan }^{2}45^{\circ }+{\cos }^{2}30^{\circ }-{\sin }^{2}90^{\circ }=\frac{7}{4}$ (that is, $1.75$).

   Step-by-step solution →
6. Q166 CBSE 2025
   If $a\sec \theta +b\tan \theta =m$ ​and $b\sec \theta +a\tan \theta =n$, ​prove that $a^2+n^2=b^2+m^2$.
   Show answer

   Squaring the given relations and subtracting, with ${\sec }^2\theta -{\tan }^2\theta =1$, gives $m^2-n^2=a^2-b^2$, that is, $a^2+n^2=b^2+m^2$. Hence proved.

   Step-by-step solution →
7. Q167 CBSE 2025
   Use the ​identity ${\sin }^{2}A+{\cos }^{2}A=1$ to ​prove that ${\tan }^{2}A+1={\sec }^{2}A$. Hence, find the value ​of $\tan A$, when $\sec A=\frac{5}{3}$, where $A$ ​is ​an ​acute angle.
   Show answer

   Dividing ${\sin }^{2}A+{\cos }^{2}A=1$ by ${\cos }^{2}A$ proves ${\tan }^{2}A+1={\sec }^{2}A$, and with $\sec A=\frac{5}{3}$ it gives $\tan A=\frac{4}{3}$.

   Step-by-step solution →
8. Q168 CBSE 2025
   Verify ​that $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$ ​for $A=30^{\circ }$.
   Show answer

   LHS $=\cos 60^{\circ }=\frac{1}{2}$ and RHS $=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2}$, so LHS $=$ RHS and the identity is verified for $A=30^{\circ }$.

   Step-by-step solution →
9. Q169 CBSE 2025
   If $\tan A=\sqrt{3}$, ​where ​A is ​an acute angle, ​then find ​the ​value of $\frac{{\sin }^{2}A}{1+{\cos }^{2}A}$.
   Show answer

   The value of $\frac{{\sin }^{2}A}{1+{\cos }^{2}A}$ is $\frac{3}{5}$.

   Step-by-step solution →
10. Q170 CBSE 2025
    It ​is given ​that $\sin (A-B)=\sin A\cos B-\cos A\sin B$. Use it ​to ​find ​the ​value ​of $\sin 15^{\circ }$.
    Show answer

    Using $\sin 15^{\circ }=\sin (45^{\circ }-30^{\circ })$, the value is $\sin 15^{\circ }=\frac{\sqrt{6}-\sqrt{2}}{4}$ (equivalently $\frac{\sqrt{3}-1}{2\sqrt{2}}$), which is about $0.2588$.

    Step-by-step solution →
11. Q171 CBSE 2025
    If $\sin A=y$, ​then ​express $\cos A$ ​and $\tan A$ in terms of $y$.
    Show answer

    In terms of $y$, $\cos A=\sqrt{1-y^2}$ and $\tan A=\frac{y}{\sqrt{1-y^2}}$.

    Step-by-step solution →
12. Q172 CBSE 2025
    Evaluate ​: $\frac{\cos 45^{\circ }}{\sec 30^{\circ }+\csc 30^{\circ }}$.
    Show answer

    $\frac{\cos 45^{\circ }}{\sec 30^{\circ }+\mathrm{cosec}30^{\circ }}=\frac{3\sqrt{2}-\sqrt{6}}{8}$ (equivalently $\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}$)

    Step-by-step solution →
13. Q173 CBSE 2025
    Find ​the ​values ​of $A$ ​and $B$ ($0\le A<90^{\circ }$, $0\le B<90^{\circ }$), ​if $\tan (A+B)=1$ and $\tan (A-B)=\frac{1}{\sqrt{3}}$.
    Show answer

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

    Step-by-step solution →
14. Q174 CBSE 2025
    Prove ​that $(\csc \theta +\sin \theta )(\csc \theta -\sin \theta )={\cot }^2\theta +{\cos }^2\theta$.
    Show answer

    Writing $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$ and using $1-{\sin }^2\theta ={\cos }^2\theta$, the left side reduces to ${\cot }^2\theta +{\cos }^2\theta$, so $\text{LHS}=\text{RHS}$. Hence proved.

    Step-by-step solution →
15. Q175 CBSE 2025
    If $\sin 3A=1$, ​then find ​the ​value ​of $\cos 2A-{\tan }^{2}45^{\circ }$.
    Show answer

    $\cos 2A-{\tan }^{2}45^{\circ }=-\frac{1}{2}$ (with $A=30^{\circ }$).

    Step-by-step solution →
16. Q176 CBSE 2025
    Find ​the ​value ​of $x$ for ​which $(\sin A+\text{cosec }A{)}^2+(\cos A+\sec A{)}^2=x+{\tan }^{2}A+{\cot }^{2}A$
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    The value of $x$ is $7$.

    Step-by-step solution →
17. Q177 CBSE 2025
    If $\sin (A+2B)=\frac{\sqrt{3}}{2}$ ​and $\cos (A+4B)=0$, $A>B$, ​then find ​the ​value ​of $A$ and $B$.
    Show answer

    $A=30^{\circ }$ and $B=15^{\circ }$

    Step-by-step solution →
18. Q178 CBSE 2025
    Evaluate: $\frac{\cos 45^{\circ }}{\tan 30^{\circ }+\sin 60^{\circ }}$.
    Show answer

    $\frac{\cos 45^{\circ }}{\tan 30^{\circ }+\sin 60^{\circ }}=\frac{\sqrt{6}}{5}$

    Step-by-step solution →
19. Q179 CBSE 2025
    Evaluate ​the ​following: $\frac{3\sin 30^{\circ }-4{\sin }^{3}30^{\circ }}{2{\sin }^{2}50^{\circ }+2{\cos }^{2}50^{\circ }}$
    Show answer

    The value of the expression is $\frac{1}{2}$.

    Step-by-step solution →
20. Q180 CBSE 2025
    Prove ​that $\tan 45^{\circ }=1$ geometrically.
    Show answer

    In right $△ABC$ with $\angle B=90^{\circ }$ and $\angle C=45^{\circ }$, the angle sum gives $\angle A=45^{\circ }$, so $AB=BC=x$ and $\tan 45^{\circ }=\tan C=\frac{AB}{BC}=\frac{x}{x}=1$. Hence proved.

    Step-by-step solution →