physicscatalyst.com logo





Class 10 trigonometry Problems with Solutions




Question 1.
Given
$a cosec A = p$ and $b cot A = q$,
then prove that
$\frac {p^2}{a^2} - \frac {q^2}{b^2} =1$

Answer

$cosec A = \frac {p}{a}$

$cosec^2 A = \frac {p^2}{a^2}$

Similarly

$ cot A = \frac {q}{b}$
$cot^2 A = \frac {q^2}{b^2}$
Now
$cosec^2 A - cot^2 A=1$
$\frac {p^2}{a^2} - \frac {q^2}{b^2} =1$



Question 2.
Prove the following trigonometric identities:
a.$cos^2 \theta (1 + tan^2 \theta) = 1$
b.$cos^2 \theta + \frac {1}{1+ cot^2 \theta} =1$
c.$ \frac {1}{1+ sin \theta } + \frac {1}{1-sin \theta } = 2 sec^2 \theta$
d.$cosec^2 \theta + sec^2 \theta =cosec^2 \theta sec^2 \theta$
e.$tan \theta - cot \theta = \frac {2sin^2 \theta -1}{sin \theta cos \theta}$

Question 3.
If A, B, C are interior angles of $\Delta ABC$, show that
$cosec ^2 (\frac {B+C}{2} ) - tan ^2 \frac {A}{2} =1$

Question 4.
If $x = a sec \theta + b tan \theta $ and $y=a tan \theta + b sec \theta $ ,prove that
$x^2 -y^2 = a^2 -b^2$

Answer

$x = a sec \theta + b tan \theta $
$x^2 =a^2 sec^2 \theta + b^2 tan^2 \theta + ab sec \theta tan \theta$
Similarly
$y=a tan \theta + b sec \theta $
$y^2 = a^2 tan^2 \theta + b^2 sec^2 \theta + ab sec \theta tan \theta$
Now
$x^ -y^2$
$=(a^2 -b^2)(sec^2 \theta -tan^2 \theta)$
$=(a^2 -b^2)$



Question 5.
Prove the following identities:
i. $2(sin^6 \theta + cos^6 \theta ) -3 (sin^4 \theta + cos^4 \theta ) + 1=0$
ii. $sin^6 \theta + cos^6 \theta + 2 sin^2 \theta cos ^2 theta =1$
iii. $(sin^8 \theta -cos^8 \theta = (sin^2 \theta -cos^2 \theta) (1 -2sin^2 \theta cos^2 \theta )$



Question 6.
If $cosec A -cot A = q$, then show that
$\frac {q^2 -1}{q^2 + 1} + cos A =0$

Answer

$cosec A -cot A = q$
$\frac {1}{sin A} - \frac {cos A}{sin A} =q$
$ \frac {1-cos A }{sin A}=q$
$ \frac {(1-cos A)^2}{sin^2 A} = q^2$
$\frac {1- cos A}{1+ cos A} = q^2$
$ \frac {q^2 -1}{q^2 + 1} + cos A =0$



Question 7.
If $x = p sec \alpha cos \beta$, $y = q sec \alpha sin \beta$ and $z = r tan \alpha$, then show that
$\frac {x^2}{p^2} + \frac {y^2}{q^2} -\frac {z^2}{r^2} =1$

Answer

$x = p sec \alpha cos \beta$
$\frac {x}{p} = sec \alpha cos \beta$
Squaring
$\frac {x^2}{p^2}= sec^2 \alpha cos^2 \beta$ -(1)

$y = q sec \alpha sin \beta$
$\frac {y}{q} =sec \alpha sin \beta$
Squaring
$\frac {y^2}{q^2} = sec^2 \alpha sin^ \beta$ --(2)
Adding (1) and (2)
$\frac {x^2}{p^2} + \frac {y^2}{q^2} =sec^2 \alpha (cos^2 \beta + sin^2 \beta)$
$\frac {x^2}{p^2} + \frac {y^2}{q^2} =sec^2 \alpha$

$z = r tan \alpha$
$ \frac {z}{r} = tan \alpha$
Squaring
$\frac {z^2}{r^2}= tan^2 \alpha$

Now
$sec^2 \alpha - tan^2 \alpha = 1$
$\frac {x^2}{p^2} + \frac {y^2}{q^2} -\frac {z^2}{r^2} =1$



Question 8.
Prove that
$(1+ cot \theta) (1 + tan \theta + sec \theta )=2$

Question 9.
Prove that
cot4 A – 1 = cosec4 A – 2cosec2 A
Question 10.
$\frac {sin A -cos A +1}{sin A + cos A -1} = \frac {1}{sec A -tan A}$
Question 11.
if $tan A + sin A =m$ and $tan A -sin A =n$
Prove that
$m^ -n^2 = 4 \sqrt {mn}$

Answer

$tan A + sin A =m$ --(1)
$tan A -sin A =n$ --(2)
Squaring and subtracting
$m^2 -n^2 = tan^2 A + sin^2 A +2 tan A sin A - tan^2 A -sin^2 A + 2tan A sin A$
$m^2 -n^2 = 4tan A sin A$

Multiplying 1 and 2
$tan^2 A -sin^2 A = mn$
$sin^2 A ( \frac {1}{cos^2 A} -1) = mn$
$sin^2 A \frac {1-cos^2 A}{cos^2 A}=mn$
$ tan^2 A (1-cos^2A) =mn$
$tan^2 A sin^2 A =mn$
$tan A sin A = \sqrt {mn}$
Hence
$m^ -n^2 = 4 \sqrt {mn}$



Question 12
$sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A$

Answer

LHS
$sin A(1 + tan A) + cos A(1 + cot A)$
$= sin A( 1 + \frac {sin A}{cos A}) + cos A ( 1 + \frac {cos A }{sin A})$
$= sin A \frac {sin A + cos A}{cos A} + cos A \frac {sinA + cos A}{sin A}$
$= (sin A + cos A) ( \frac {sin A }{cos A} + \frac {cos A}{sin A})$
$= (sin A + cos A) (\frac {sin^2 A + cos^2 A}{sin A cos A})$
$=\frac {sin A + cos A}{sin A cos A}$
$=sec A + cosec A$



Question 13
$\frac {tan A}{1- cot A} + \frac {cot A}{1-tan A} = 1 + tan A + cot A$
Question 14
$\sqrt {sec^2 A + cosec^2 A} = (tan A + cot A)$

Answer

LHS
$\sqrt {sec^2 A + cosec^2 A}$
$= \sqrt {1 + tan^2 A + 1 + cot^2 A}$
$= \sqrt { tan^2 A + cot^2 A + 2}$
$= \sqrt { tan^2 A + cot^2 A + 2 tan A cot A}$ as tanA cot A=1
$= \sqrt {(tan A + cot A)^2}$
$=tan A + cot A$



Question 15
if $xsin^3 A + ycos^3 A = sin A cos A$
and $x sin A = y cos A$
Prove that
$x^2 + y^2 =1$

Answer

$xsin^3 A + ycos^3 A = sin A cos A$
$(xsin A) sin^2A + (ycosA)cos^2 A = sin A cos A$
Now $x sin A = y cos A$
$(xsin A) sin^2A + (x sin A)cos^2 A = sin A cos A$
$xsinA (sin^2 A + cos^2 A) =sin A cos A$
$x=cos A$
Now
$x sin A = y cos A$
$cos A sin A = y cos A$
$y=sin A$
Now
$x^2 + y^2 = sin^2 A + cos^2 A =1$


Summary

This Class 10 trigonometry Problems 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


Download Trigonometry Problems as pdf
Also Read




Go back to Class 10 Main Page using below links
Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20



Latest Updates
Synthetic Fibres and Plastics Class 8 Practice questions

Class 8 science chapter 5 extra questions and Answers

Mass Calculator

3 Fraction calculator

Garbage in Garbage out Extra Questions7