Sample QuestionsIntroduction to Trigonometry questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
$\frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A }=$
- A
$sec^2A$
- B
$–1$
- C
$cot^2A$
- ✓
$tan^2A$
Answer: D.
View full solution →$(sec A + tan A) (1 – sin A)$
- A
$sec A$
- B
$sin A$
- C
$cosec A$
- ✓
$cos A$
Answer: D.
View full solution →($1$ + tan $\theta$ + sec $\theta$) ($1$ + cot$\theta$ – cosec$\theta$) =
Answer: C.
View full solution →$9 sec^2 A – 9$ $ tan^2 A =$
Answer: A.
View full solution →$\tan ^2 \theta-\sec ^2 \theta=$_______ $(1,-1,0)$
View full solution →If $\tan ^2 \theta=\sin ^2 \theta+\cos ^2 \theta$ then _______ $\left(45^{\circ}, 1^{\circ}, 90^{\circ}\right)$
View full solution →If point $P \left(x_1, 0\right)$ and $Q \left(x_2, 0\right)$ are two points on the $X$-axis then distance between them $PQ$_______ $\left(\left|x_1-0\right| \cdot\left|x_2-0\right|,\left|x_1-x_2\right|\right)$
View full solution →$\tan ^2 \theta-\sec ^2 \theta=$ _______ $\quad(0<0<90) .(1,-1,0)$
View full solution →$\tan 45^{\circ}+\sin 90^{\circ}-\sec 60^{\circ}=$_______ . $\quad(0,10,2)$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$(cosec A - sin A) (sec A - cos A) =$ $\frac { 1 } { \tan A + \cot A }$
[Hint: Simplify $LHS$ and $RHS$ separately]
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$\sqrt { \frac { 1 + \sin A } { 1 - \sin A } }$ $= sec A + tan A$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\frac { 1 + \sec A } { \sec A } = \frac { \sin ^ { 2 } A } { 1 - \cos A }$
[Hint: Simplify $LHS$ and $RHS$ separately]
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$\left( \frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A } \right) = \left( \frac { 1 - \tan A } { 1 - \cot A } \right) ^ { 2 } = tan^2 A$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined.
$( cosec\; \theta - \cot \theta ) ^ { 2 } = \frac { 1 - \cos \theta } { 1 + \cos \theta }$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\frac { \sin \theta - 2 \sin ^ { 3 } \theta } { 2 \cos ^ { 2 } \theta - \cos \theta } = \tan \theta$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined. $\frac { \cos A } { 1 + \sin A } + \frac { 1 + \sin A } { \cos A } = 2 \sec A$
View full solution →If $tan\ (A + B) =$ $\sqrt3$ and $tan\ (A - B) =$ $\frac{1}{\sqrt3}$; $0^\circ < A + B$ $\leq$ $90^\circ ; A > B$, then find $A$ and $B$.
View full solution →Evaluate: $\frac { 5 \cos ^ { 2 } 60 ^ { \circ } + 4 \sec ^ { 2 } 30 ^ { \circ } - \tan ^ { 2 } 45 ^ { \circ } } { \sin ^ { 2 } 30 ^ { \circ } + \cos ^ { 2 } 30 ^ { \circ } }$
View full solution →Evaluate: $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ cosec 60^{\circ} }{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
View full solution →Prove the given identities, where the angles involved are acute angles for which the expressions are defined. $(\sin A+\operatorname{cosec} A)^2+(\cos A+\sec A)^2=7+\tan ^2 A+\cot ^2 A$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined.$\frac{\cos A-\sin A+1}{\cos A+\sin A-1} = cosec A + \cot A,$ using the identity $\operatorname{cosec}^2 \mathrm{~A}=1+\cot ^2 \mathrm{~A}$
View full solution →Prove the given identity, where the angles involved are acute angles for which the expressions are defined.$ \frac{{\tan A }}{{1 - \cot A }} + \frac{{\cot A }}{{1 - \tan A }} = 1 + sec A \cos ecA$
[Hint: Write the expression in terms of $\sin \theta$ and$ \cos \theta$]
View full solution →Write all the other trigonometric ratios of $\angle$A in terms of $sec A.$
View full solution →Express the trigonometric ratios $\sin A , \sec A$ and $\tan A$ in terms of $\cot A$.
View full solution →
| Section-A |
Section- B |
| $Q.1. \sec ^2 \theta-\tan ^2 \theta$ |
$(a)$ |
$\operatorname{cosec} \theta$ |
| $Q.2. \frac{1}{\sin \theta}$ |
$(b)$ |
$\cos ^2 \theta$ |
| |
$(c)$ |
$-1$ |
| Section-A |
Section-B |
| $Q.1. \sin \theta$ |
$(a)$ |
$-\cot ^2 \theta$ |
| $Q.2. \cos \theta$ |
$(b)$ |
$\sqrt{1-\cos ^2 \theta}$ |
| |
$(c)$ |
$\sqrt{1-\sin ^2 \theta}$ |
If $\sin \theta=\frac{3}{5}$ then match $A$ with $B$
| Section-A |
Section-B |
| $Q.1. \operatorname{cosec} \theta$ |
$(a)$ |
$\frac{4}{5}$ |
| $Q.2. \sec \theta$ |
$(b)$ |
$\frac{5}{4}$ |
| |
$(c)$ |
$\frac{5}{3}$ |
View full solution →$\sin \theta=\frac{3}{5}$ then match $A$ with $B$
| Section-A |
Section-B |
| $Q.1. \cos \theta$ |
$(a)$ |
$\frac{3}{4}$ |
| $Q.2. \tan \theta$ |
$(b)$ |
$\frac{4}{5}$ |
| |
$(c)$ |
$\frac{5}{4}$ |
View full solution →
| Section-A |
Section-B |
| $Q.1. \cot ^2 \theta-\operatorname{cosec}^2 \theta$ |
$(a)$ |
$\operatorname{cosec} \theta$ |
| $Q.2. 1-\sin ^2 \theta$ |
$(b)$ |
$\cos ^2 \theta$ |
| |
$(c)$ |
$-1$ |