Trigonometry Important Questions for Class 10 Maths
Given below are the Class 10 Maths trigonometry extra questions.This includes both important and tough questions
a. Multiple Choice Questions
b. Short Answer type
c. Long answer type
d. Fill in the blanks
e. True and false
Multiple Choice Questions
Question 1
If $x(cos A) - y(sin A) = a$, $x(sin A) + y(cos A) = b$, the tick mark whichever option is correct
a. $x^2 -y^2 = a^2 -b^2$
b. $x^2 +y^2 = a^2 +b^2$
c. $x^2 +y^2 = a^2 - b^2$
d. $x^2 -y^2 = a^2 + b^2$ Solution
$x(cos A) - y(sin A) = a$, $x(sin A) + y(cos A) = b$
Squaring both the equation and adding
$x^2(cos^2 A + sin^2 A) + y^2(cos^2 A + sin^2 A) -2xy sin A cos A + 2xy sin A cos A=a^2 + b^2$
Now as $ sin^2 A + cos^2 A =1$
$x^2 + y^2 =a^2 + b^2$
Answer (b)
Question 2
If $cos A=\frac {1}{2}$, $sin B =\frac {1}{2}$ then value of A +B
a. 30°
b. 60°
c. 90°
d. 120° Solution
$cos A=\frac {1}{2}$
$cos A = cos 60$
A=60°
$sin B =\frac {1}{2}$
$sin B = sin 30$
B=30°
Therefore,
A+ B= 90°
Answer (c)
Question 3
If $sin (X + Y) = cos (X - Y) =1$ then
a. X = Y = 90°
b. X = Y = 0°
c. X = Y = 45°
d. X = 2Y Solution
$sin (X + Y) =1$
$sin (X+ Y) = sin90$
X+ Y=90
$cos (X - Y) =1$
$cos( X-Y) =cos 0$
X-Y =0
Solving these we get
X = Y = 45°
Answer (c)
Short Answer type
Question 4.
If $sec X= a + \frac {1}{4a}$, prove that $sec X+ tan X=2a \; or \; \frac {1}{2a}$ Solution
For positive sign
Now $Sec X + Tan X = (a + \frac {1}{4a}) + (a - \frac {1}{4a})$
= 2a
For negative sign
Now $Sec X + Tan X = (a + \frac {1}{4a}) - (a - \frac {1}{4a})$
$ = \frac {1}{2a}$
Question 5.
If $Sin A+ Sin^2 A= 1$, then find the value of $(cos^2 A+cos^4 A)$. Solution
$Sin A+ Sin^2 A= 1$
$sin A = 1 -sin^2 A$
$sin A =cos^2 A$
Now
$(cos^2 A+cos^4 A)$
$=(sinA + sin^2 A)$
$=1$
Question 6
If $tan 2A = cot (A - 18^{\circ})$, where 2A is an acute angle. Find the value of A. Solution
As Cot ( 90 - x ) = tan x
According to the problem given
$tan 2A = cot (A - 18^{\circ})$
$Cot ( 90 - 2A ) = cot ( A - 18^{\circ})$
or
90 - 2A = A - 18
3A =108
A=36 °
Question 7
If $tan (A+B) = \sqrt {3}$ and $tan (A - B) = \frac {1}{\sqrt {3}}$ Find the value of A and B. Solution
Question 11
If A and B acute angles such that $tan A = \frac {1}{2}$ , $tan B = \frac {1}{3}$ and
$tan (A + B) =\frac { tan A + tan B}{1- tan A tan B}$, find A + B. Solution
$tan (A + B) =\frac { tan A + tan B}{1- tan A tan B}$
$=\frac { 1/2 + 1/3}{1 - 1/2 \times 1/3} = 1$
$tan (A + B) =tan 45$
A + B =45
Question 12
Prove that
a. $tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ} = 1$
b. $tan 1^{\circ} tan 2^{\circ} tan 3^{\circ} .... tan 89^{\circ} = 1$
c. $cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ} = 0$ Solution
a. $tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ} = 1$
LHS
$tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ}$
$=tan 10^{\circ} tan 15^{\circ} tan (90^{\circ} -15 ^{\circ}) tan (90^{\circ} -10^{\circ})$
$=tan 10^{\circ} tan 15^{\circ} cot 15^{\circ} cot 10^{\circ}$
$=tan 10^{\circ} tan 15^{\circ} \frac {1}{tan 15^{\circ}} \frac {1}{tan 10^{\circ}}$
$=1$
c. $cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ}$
LHS
Since cos 90=1
$cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ}=1$
Question 13
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
a. $ \frac {cos A}{1 + sin A} + \frac {1 + sin A}{cos A} = 2 sec A$
b. $ \frac {1 + sec A}{sec A} = \frac {sin^2 A}{1- cos A}$
c. $ \sqrt { \frac {1+ sin A}{1- sin A} } = sec A + tan A$ Solution
a. LHS
$ \frac {cos A}{1 + sin A} + \frac {1 + sin A}{cos A}$
$= \frac {cos^2 A + 1 +sin^2 A + 2 sinA}{cosA(1+ SinA)}$
$=\frac {2+ 2 sin A}{cos A(1+sin A)}$
$= \frac {2}{cos A}$
$=2 sec A$
=RHS
RHS
$\frac {sin^2 A}{1- cos A}$
$= \frac {1 - cos^2 A}{1- cos A}$
$=1 + cos A$
c. LHS
$ \sqrt { \frac {1+ sin A}{1- sin A} }$
$=\sqrt { \frac {1+ sin A}{1- sin A} \times \frac {1+ sin A}{1 +sin A} }$
$=\sqrt { \frac {(1+sin A)^2}{cos^2 A}}$
$= \frac {1+ sin A}{cos A}$
$=sec A + tan A$
Question 14
In a $\Delta ABC$ right angled at C, if $tan A = \frac {1}{\sqrt {3}}$ find the value of
$sin A cos B + cos A sin B$. Solution
$tan A = \frac {1}{\sqrt {3}}$
$tan A = tan 30$
A=30°
Now C=90°
B=60°
Then
$sin A cos B + cos A sin B$
$=sin 30 cos 60 + cos 30 sin 60$
$ =\frac {1}{4} + \frac {3}{4}$
$=1$
Question 15
If $ sec \theta - tan \theta = x$, show that:
$sec \theta = \frac {1}{2} [x + \frac {1}{x}]$
$tan \theta =\frac {1}{2} [\frac {1}{x} -x]$ Solution
$ sec \theta - tan \theta = x$ - (1)
$(sec \theta - tan \theta) \times \frac {sec \theta - tan \theta}{sec \theta + tan \theta} = x$
$\frac {sec^2 \theta - tan^2 \theta}{sec \theta + tan \theta}=x$
$\frac {1}{sec \theta + tan \theta}=x$
$sec \theta + tan \theta = \frac {1}{x}$ -(2)
Adding (1) and (2)
$2 sec \theta = x + \frac {1}{x}$
$sec sec \theta=\frac {1}{2} [x + \frac {1}{x}]$
subtracting (1) from (2)
$2 tan \theta= \frac {1}{x} -x$
$tan \theta =\frac {1}{2} [\frac {1}{x} -x]$
Question 16.
If $tan \theta = \frac {12}{5}$
Find the value $ \frac {1+ sin \theta }{1 -sin \theta }$ Solution
$tan \theta = \frac {12}{5}$
$\frac {P}{B} =\frac {12}{5}$
So P=12k and B=5k
$H^2 =P^2 + B^2= (13k)^2$
H=13K
$ sin \theta = \frac {12}{13}$
Now
$ \frac {1+ sin \theta }{1 -sin \theta }$
$= \frac {25}{1} = 25$
Question 17
If $sin \theta + cos \theta = \sqrt {2} cos (90 - \theta)$,find $cot \theta$ Solution
$sin \theta + cos \theta = \sqrt {2} cos (90 - \theta)$
$sin \theta + cos \theta = \sqrt {2} sin \theta$
$cos \theta= sin \theta ( \sqrt {2} -1)$
$cot \theta=\sqrt {2} -1$
Question 18 Prove that
If $ tan^2 \theta = 1 -p^2 $, then prove that $sec \theta + tan^3 \theta cosec \theta = (2 - p^2) ^ {3/2}$. Solution
Question 19
(i) If sin A =4/5, the value of cos A = _____
(ii) if tan A + cot A =2, then the value of $tan^2 A +cot^2 A =$ ____
(iii) if cos A = 12/13, then the value of tan A = _____
(iv) The value $sin ^2 30 + sin^2 60 =$ _______ Solution
(i) 3/5
(ii)2
(ii) 5/12
(iv) 2
True and false
Question 20
(i) $cos^4 A - sin^4 A + 1=2 sin^2 A$
(ii)$\frac {tan 46^0}{cot 44^0} =1$
(iii)$tan^2 A + cot^2 A=1$
(iv) The value of cos A is $p + \frac {1}{p}$, where p is a positive number.
Solution
(i)False
(ii)True
(ii)False
(iv) false
Summary
This Class 10 trigonometry Extra Questions with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.You can download in PDF form also using the below links
