Introduction to Trigonometry: Class 10 PYQs 2018–2026 with solutions

When $\sin A=\frac{1}{3}$, the value of $\cot A$ is

$\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}$ equals to:

For an acute angle $\theta$, if $\sin \theta =\frac{1}{9}$, then value of $\frac{9\csc \theta +1}{9\csc \theta -1}$ is

Simplest form of $\frac{\sec A}{\sqrt{{\sec }^{2}A-1}}$ is

If $\theta$ is an acute angle and $7+4\sin \theta =9$, then the value of $\theta$ is:

The value of ${\tan }^2\theta -\left(\frac{1}{\cos \theta }\times \sec \theta \right)$ is:

If $7{\cos }^2\theta +3{\sin }^2\theta =4$, then the value of $\theta$ is:

If $\sin 30^{\circ }\tan 45^{\circ }=\frac{\sec 60^{\circ }}{k}$, then the value of $k$ is:

If $\sin \theta =\cos \theta$, $(0^{\circ }<\theta <90^{\circ })$, then value of $(\sec \theta \cdot \sin \theta )$ is:

Assertion (A): If $\sin A=\frac{1}{3}$ $(0^{\circ }<A<90^{\circ })$, then the value of $\cos A$ is $\frac{2\sqrt{2}}{3}$.
Reason (R): For every angle $\theta$, ${\sin }^2\theta +{\cos }^2\theta =1$.

If $\sin \theta =1$, then the value of $\frac{1}{2}\sin \left(\frac{\theta }{2}\right)$ is:

$\sec \theta$ when expressed in terms of $\cot \theta$, is equal to:

The value of $\left({\sin }^2\theta +\frac{1}{1+{\tan }^2\theta }\right)=$ ______.
OR
The value of $(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )=$ ______.

If $\cot \theta =\frac{7}{8}$, then the value of $\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=$ ______.

What is the value of $\left(\frac{1}{1+{\cot }^2\theta }+\frac{1}{1+{\tan }^2\theta }\right)$?

Evaluate: ${\sin }^{2}60^{\circ }+2\tan 45^{\circ }-{\cos }^{2}30^{\circ }$

If $\sin A=\frac{3}{4}$, calculate $\sec A$.

If $\sin x+\cos y=1$; $x=30^{\circ }$ and $y$ is an acute angle, find the value of $y$.

Prove that: $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

Evaluate: $\frac{3{\cos }^{2}30^{\circ }-6{\csc }^{2}30^{\circ }}{{\tan }^{2}60^{\circ }}$

Evaluate: $\frac{{\sin }^{3}60^{\circ }-\tan 30^{\circ }}{{\cos }^{2}45^{\circ }}$

For acute angles A and B and $A+2B$ and $2A+B$ are acute if $\tan (A+2B)=\sqrt{3}$ and $\sin (2A+B)=\frac{1}{\sqrt{2}}$, then find the measures of angles A and B.

If $x\cos 60^{\circ }+y\cos 0^{\circ }+\sin 30^{\circ }-\cot 45^{\circ }=5$, then find the value of $x+2y$.

Evaluate: $\frac{{\tan }^{2}60^{\circ }}{{\sin }^{2}60^{\circ }+{\cos }^{2}30^{\circ }}$

If $\tan A=\sqrt{3}$; where A is an acute angle, then find the value of $\frac{{\sin }^{2}A}{1+{\cos }^{2}A}$.

Evaluate: $2{\sin }^{2}30^{\circ }\sec 60^{\circ }+{\tan }^{2}60^{\circ }$.

If $2\sin (A+B)=\sqrt{3}$ and $\cos (A-B)=1$, then find the measures of angles A and B. $0\le A,B,(A+B)\le 90^{\circ }$.

Evaluate: $\frac{5{\cos }^{2}60^{\circ }+4{\sec }^{2}30^{\circ }-{\tan }^{2}45^{\circ }}{{\sin }^{2}30^{\circ }+{\sin }^{2}60^{\circ }}$

If $\sin (A-B)=\frac{1}{2}$, $\cos (A+B)=\frac{1}{2}$; $0<A+B\le 90^{\circ }$, $A>B$; find $\angle A$ and $\angle B$.

If $4{\cot }^{2}45^{\circ }-{\sec }^{2}60^{\circ }+{\sin }^{2}60^{\circ }+p=\frac{3}{4}$, then find the value of $p$.

If $\cos A+{\cos }^{2}A=1$, then find the value of ${\sin }^{2}A+{\sin }^{4}A$.

Evaluate $\frac{5{\cos }^{2}60^{\circ }+4{\sec }^{2}30^{\circ }-{\tan }^{2}45^{\circ }}{{\sin }^{2}30^{\circ }+{\cos }^{2}30^{\circ }}$

If A and B are acute angles such that $\sin (A-B)=0$ and $2\cos (A+B)-1=0$, then find angles A and B.

The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is supporting the disc as shown in Fig. 4. If $AC=1.5$ m long and $CD=3$ m, find (i) $\tan \theta$ (ii) $\sec \theta +\csc \theta$

If $\tan \theta =\frac{3}{4}$, find the value of $\left(\frac{1-{\cos }^2\theta }{1+{\cos }^2\theta }\right)$

If $\tan \theta =\sqrt{3}$, find the value of $\left(\frac{2\sec \theta }{1+{\tan }^2\theta }\right)$

Prove that: $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$

Prove that: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$

Prove that: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$

Prove that: $\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{2}{2{\sin }^{2}A-1}$

Prove that: $\left(1+\frac{1}{{\tan }^2\theta }\right)\left(1+\frac{1}{{\cot }^2\theta }\right)=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$

Prove that: $\sqrt{\frac{\csc \theta -1}{\csc \theta +1}}+\sqrt{\frac{\csc \theta +1}{\csc \theta -1}}=2\sec \theta$

Prove that $\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }=\frac{1-\sin \theta }{\cos \theta }$.

Prove that $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$

Prove that: $\left(\frac{1}{\cos \theta }-\cos \theta \right)\left(\frac{1}{\sin \theta }-\sin \theta \right)=\frac{1}{\tan \theta +\cot \theta }$.

Prove that $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$

Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$.

If $\sin \theta +\cos \theta =\sqrt{3}$, then prove that $\tan \theta +\cot \theta =1$.

If $\sin \theta +\cos \theta =p$ and $\sec \theta +\csc \theta =q$, show that $q(p^2-1)=2p$.

Prove that: $\frac{\tan \theta }{1-\tan \theta }-\frac{\cot \theta }{1-\cot \theta }=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$

If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta$, show that $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$.

If $4\tan \theta =3$, evaluate $\left(\frac{4\sin \theta -\cos \theta +1}{4\sin \theta +\cos \theta -1}\right)$

Prove that: $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$

Prove that: $\frac{\sin \theta }{\cot \theta +\csc \theta }=2+\frac{\sin \theta }{\cot \theta -\csc \theta }$

Prove that: $\frac{(1+\cot \theta +\tan \theta )(\sin \theta -\cos \theta )}{({\sec }^3\theta -{\csc }^3\theta )}={\sin }^2\theta {\cos }^2\theta$

Prove that: $\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-\cos A}=\tan A$.