# ICSE Class 10 Trigonometric Identities Board PYQ with Solution

This post contains ICSE Class 10 Trigonometric Identities Board PYQ with Solution. The sums shown here have come in ICSE Board Examinations from the year 2010-2023.

Trigonometry is a very important part of Mathematics syllabus for ICSE Class 10. This is a rather new concept for students and thus many students get bewildered when they encounter sums from Trigonometry in Board Exams, especially from Trigonometric Identities. Hence, it is very important to practice this chapter properly, so that the students are confident enough to solve the sums of Trigonometic Identities.

So, it is very important to solve the PYQs (previous years’ questions) from Boards. By solving the PYQs , students will know what kind of sums from Trigonometric Identities have been asked and could be asked in Board exams.

## ICSE Class 10 Trigonometric Identities Board PYQ with Solution

1.Prove the following identity: (sin2θ -1) (tan2θ + 1) + I = 0 [2023]

Step-by-step Explanation:

$\;LHS\\=\;(\sin^2\theta\;-1)\;(\tan^2\theta\;+\;1)\;+\;I\;\;\;\;\;\;\;\;\;\\=\;(-\;\cos^2\theta)\;(sec^2\theta)\;+\;1\\=(-\;\cos^2\theta)\;\times\frac1{\cos^2\theta}+1\\=-1+1\\=0\\=\;RHS\\proved$

2. (1 + sin A) (1 – sin A) is equal to:
(a) cosec2A
(b) sin2A
(c) sec2A
(d) cos2A [2023]

Solution: (d)

Step-by-step Explanation:

(1 + sin A) (1 – sin A)

= 1 – sin2A

= cos2A

You can also watch video solution of Trigonometric Identities Board PYQs HERE.

$\;3.\;Prove\;that:\;1+\frac{\tan^2\theta}{1+sec\theta}=sec\theta\;\;\;\;\;\;\;\;\;\lbrack2022\;Semester-2\rbrack$

Step-by-step Explanation:

$\;LHS\\=\;1+\frac{\tan^2\theta}{1+sec\theta}\\=1+\frac{sec^2\theta-1}{sec\theta+1}\\=1+\frac{(sec\theta+1)\;(sec\theta-1)}{(sec\theta+1)}\\=1+sec\theta-1\\=sec\theta\\=\;RHS\\proved$

4. Prove that: [2022 Semester-2]

$\frac{{(1+\sin\theta)}^2+{(1-\sin\theta)}^2}{2\cos^2\theta}=sec^2\theta+\tan^2\theta$

Step-by-step Explanation:

$LHS\\=\frac{\;{(1+\sin\theta)}^2+{(1-\sin\theta)}^2}{2\cos^2\theta}\\=\frac{1+\sin^2\theta+2\sin\theta+1+\sin^2\theta-2\sin\theta}{2\cos^2\theta}\\=\frac{2+2\sin^2\theta}{2\cos^2\theta}\\=\frac{2(1+\sin^2\theta)}{2\cos^2\theta}\\=\frac{1+\sin^2\theta}{\cos^2\theta}\\=\frac1{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}\\=sec^2\theta+\tan^2\theta\\=\;RHS\\Proved.$

$5.Prove\;that:\frac1{1+sin\theta}+\frac1{1-sin\theta}=2sec^2\theta\;\;\;\;\;\;\;\;\;\lbrack2022Semester-2\rbrack$

Step-by-step Explanation:

$LHS\\=\frac1{1+sin\theta}+\frac1{1-sin\theta}\\=\frac{1-sin\theta+1+sin\theta}{(1+sin\theta)\;(1-sin\theta)}\\=\frac2{1-sin^2\theta}\\=\frac2{\cos^2\theta}\\=2sec^2\theta\\=RHS\\Proved.$

$6.\;Prove\;the\;identity:\;\left(\frac{1-\tan\theta}{1-cot\theta}\right)^2=tan^2\theta\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack2020\rbrack$

Step-by-step Explanation:

$LHS\\=\;\left(\frac{1-\tan\theta}{1-cot\theta}\right)^2\\=\left(\frac{1-\tan\theta}{1-{\displaystyle\frac1{\tan\theta}}}\right)^2\\=\left(\frac{1-\tan\theta}{\displaystyle\frac{\tan\theta-1}{\tan\theta}}\right)^2\\=\left(\frac{(1-\tan\theta)\times\tan\theta}{\tan\theta-1}\right)^2\\=\left(-\;\tan\theta\right)^2\\=\tan^2\theta\\=RHS\\Proved.$

ICSE Class 10 Maths PYQs with Solution All Chapters

$7.\;Prove\;that:\frac{\sin A}{1+cotA}-\frac{\cos A}{1+\tan A}=\sin A-\cos A\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack2020\rbrack$

Step-by-step Explanation:

$LHS\\=\frac{\sin A}{1+cotA}-\frac{\cos A}{1+\tan A}\\=\frac{\sin A}{1+{\displaystyle\frac{\cos A}{\sin A}}}-\frac{\cos A}{1+{\displaystyle\frac{\sin A}{\cos A}}}\\=\frac{\sin A}{\frac{SinA+\cos A}{\sin A}}-\frac{\cos A}{\displaystyle\frac{\cos A+\sin A}{\cos A}}\\=\frac{\sin^2A}{\sin A+\cos A}-\frac{\cos^2A}{\sin A+\cos A}\\=\frac{\sin^2A-\cos^2A}{\sin A+\cos A}\\=\frac{(\sin A+\cos A)(\sin A-\cos A)}{\sin A+\cos A}\\=\sin A-\cos A\\=\;RHS\\Proved.$

8. Prove that: (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 [2019]

Step-by-step Explanation:

$LHS\\=(\cos ec\;\theta\;–\;\sin\;\theta)\;(sec\;\theta\;–\;\cos\;\theta)\;(\tan\;\theta\;+\;cot\;\theta)\\=\left(\frac1{\sin\;\theta}-\sin\;\theta\right)\;\left(\frac1{\cos\;\theta}-\cos\;\theta\right)\;\left(\frac{\sin\;\theta}{\cos\;\theta}+\frac{\cos\;\theta}{\sin\;\theta}\right)\\=\left(\frac{1-\sin^2\theta}{\sin\;\theta}\right)\left(\frac{1-\cos^2\theta}{\cos\;\theta}\right)\left(\frac{\sin^2\theta+\cos^2\theta}{\sin\;\theta.\cos\;\theta}\right)\\=\left(\frac{\cos^2\theta}{\sin\;\theta}\right)\left(\frac{\sin^2\theta}{\cos\;\theta}\right)\left(\frac1{\sin\;\theta.\cos\;\theta}\right)\\=1\\=RHS\\Proved.$

$9.\;Prove\;that:\;\sqrt{sec^2\theta+\cos ec^2\theta}=tan\theta+cot\theta\;\;\lbrack3\rbrack\;\;\;\lbrack2018\rbrack\\Step-by-step\;Explanation:\\LHS\\=\sqrt{sec^2\theta+\cos ec^2\theta}\\=\sqrt{\frac1{\cos^2\theta}+\frac1{\sin^2\theta}}\\=\sqrt{\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta.\;\sin^2\theta}}\\=\sqrt{\frac1{\cos^2\theta.\;\sin^2\theta}}\\=\frac1{\cos\theta\;\sin\theta}\\=\frac{\sin^2\theta+\cos^2\theta}{\cos\theta\;\sin\theta}\\=\frac{\sin^2\theta}{\cos\theta\;\sin\theta}+\frac{\cos^2\theta}{\cos\theta\;\sin\theta}\\=\frac{\sin\theta}{\cos\theta\;}+\frac{\cos\theta}{\sin\theta}\\=\tan\theta+cot\theta\\=\;RHS\\Proved.\\$

10. Prove that (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2 [4] [2018]

Step-by-step Explanation:

$LHS\\=(1\;+\;cot\;\theta\;–\;\cos ec\;\theta)\;(1\;+\;\tan\;\theta\;+\;sec\;\theta)\\=\left(1+\frac{\cos\theta}{\sin\theta}-\frac1{\sin\theta}\right)\;\left(1+\frac{\sin\theta}{\cos\theta}+\frac1{\cos\theta}\right)\;\\=\left(\frac{\sin\theta+\cos\theta-1}{\sin\theta}\right)\;\left(\frac{\cos\theta+\sin\theta+1}{\cos\theta}\right)\\=\left(\frac{{(\sin\theta+\cos\theta)}^2-{(1)}^2}{\sin\theta\;\cos\theta}\right)\\=\left(\frac{{\sin^2\theta+\cos^2\theta+2\;\sin\theta\;\cos\theta}-1}{\sin\theta\;\cos\theta}\right)\\=\left(\frac{1+2\;\sin\theta\;\cos\theta-1}{\sin\theta\;\cos\theta}\right)\\=\left(\frac{2\;\sin\theta\;\cos\theta}{\sin\theta\;\cos\theta}\right)\\=2\\=\;RHS\\Proved.\\=\;RHS\\Proved.\\$

$11.\;Prove\;that:\;\frac{sin\theta-2sin^3\theta}{2cos^3\theta-cos\theta}=tan\theta\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack2017\rbrack\\Step-by-step\;Explanation:\\LHS\\=\frac{sin\theta-2sin^3\theta}{2cos^3\theta-cos\theta}\\=\frac{sin\theta\;(1-2sin^2\theta)}{cos\theta\;(2cos^2\theta-1)}\\=\frac{sin\theta\;(sin^2\theta+\cos^2\theta-2sin^2\theta)}{cos\theta\;(2cos^2\theta-sin^2\theta-\cos^2\theta)}\\=\tan\theta\frac{(\cos^2\theta-sin^2\theta)}{\;(cos^2\theta-sin^2\theta)}\\=\;\tan\theta\\=RHS\\Proved.$

$12.\;Prove\;that:\;\frac{\cos A}{1+\sin A}+\tan A=secA\;\;\;\lbrack3\rbrack\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack2016\rbrack\\\\Step-by-step\;Explanation:\\\\LHS\\=\frac{\cos A}{1+\sin A}+\tan A\\=\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A}\\=\frac{\cos^2A+\sin A\;(1+\sin A)}{\cos A\;(1+\sin A)}\\=\frac{\cos^2A+\sin A+\sin^2A}{\cos A\;(1+\sin A)}\\=\frac{1+\sin A}{\cos A\;(1+\sin A)}\\=\frac1{\cos A}\\=\;secA\\=RHS\\Proved.\\$

$13.\;Prove\;that\;\frac{\sin\theta}{1-\mathrm{co}t\theta}+\frac{\cos\theta}{1-\tan\theta}=\cos\theta+\sin\theta\;\;\lbrack3\rbrack\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack2015\rbrack\\\\Step-by-step\;Explanation:\\\\LHS\\=\frac{\sin\theta}{1-\mathrm{co}t\theta}+\frac{\cos\theta}{1-\tan\theta}\\=\frac{\sin\theta}{1-{\displaystyle\frac{\cos\theta}{\sin\theta}}}+\frac{\cos\theta}{1-{\displaystyle\frac{\sin\theta}{\cos\theta}}}\\=\frac{\sin\theta}{\displaystyle\frac{\sin\theta-\cos\theta}{\sin\theta}}+\frac{\cos\theta}{\displaystyle\frac{\cos\theta-\sin\theta}{\cos\theta}}\\=\frac{\sin^2\theta}{\sin\theta-\cos\theta}+\frac{\cos^2\theta}{\displaystyle\cos\theta-\sin\theta}\\=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\displaystyle\sin\theta-\cos\theta}\\=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}\\=\frac{(\sin\theta+\cos\theta)\;(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}\\=\sin\theta+\cos\theta\\=RHS\\Proved.\\$

14. Prove the identity: (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ [3] [2014]

Step-by-step Explanation:

$13.\;Prove\;that\;\frac{\sin\theta}{1-\mathrm{co}t\theta}+\frac{\cos\theta}{1-\tan\theta}=\cos\theta+\sin\theta\;\;\lbrack3\rbrack\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack2015\rbrack\\\\Step-by-step\;Explanation:\\\\LHS\\=(\sin\;\theta\;+\;\cos\;\theta)\;(\tan\;\theta\;+\;cot\;\theta)\\=(\sin\;\theta\;+\;\cos\;\theta)\;\left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right)\\=(\sin\;\theta\;+\;\cos\;\theta)\;\left(\frac{\sin^2\theta+\cos^2\theta}{\cos\theta\;\sin\theta}\right)\\=(\sin\;\theta\;+\;\cos\;\theta)\;\left(\frac1{\cos\theta\;\sin\theta}\right)\\=\frac{\sin\theta}{\cos\theta\;\sin\theta}+\frac{\cos\theta\;}{\cos\theta\;\sin\theta}\\=\frac1{\cos\theta}+\frac{\;1}{\sin\theta}\\=sec\;\theta+\cos ec\;\theta\\=\;RHS\\Proved.\\$

$15.\;Show\;that:\;\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}\;\;\;\;\;\lbrack3\rbrack\;\;\;\;\;\;\;\;\;\;\;\lbrack2013\rbrack\\\\Step-by-step\;Explanation:\\\\LHS\\=\sqrt{\frac{1-\cos A}{1+\cos A}}\\=\sqrt{\frac{1-\cos A}{1+\cos A}\times\frac{1+\cos A}{1+\cos A}}\\=\sqrt{\frac{1-\cos^2A}{{(1+\cos A)}^2}}\\=\sqrt{\frac{\sin^2A}{{(1+\cos A)}^2}}\\=\frac{\sin A}{1+\cos A}\\=\;RHS\\Proved.$

$16.Prove\;that:\;\frac{\tan^2\theta}{{(sec\theta-1)}^2}=\frac{1+cos\theta}{1-cos\theta}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lbrack3\rbrack\;\;\;\;\;\;\lbrack2012\rbrack\\\\Step-by-step\;Explanation\\\\LHS\\=\frac{\tan^2\theta}{{(sec\theta-1)}^2}\\=\frac{sec^2\theta-1}{{(sec\theta-1)}\;(sec\theta-1)}\\=\frac{(sec\theta+1)\;(sec\theta-1)}{{(sec\theta-1)}\;(sec\theta-1)}\\=\frac{(sec\theta+1)\;}{(sec\theta-1)}\\=\frac{{\displaystyle\frac1{\cos\theta}}+1}{{\displaystyle\frac1{\cos\theta}}-1}\\=\frac{\displaystyle\frac{1+\cos\theta}{\cos\theta}}{\displaystyle\frac{1-\cos\theta}{\cos\theta}}\\=\frac{1+\cos\theta}{1-\cos\theta}\\=\;RHS\\Proved.$

17. Prove that (cosec A – sin A) (sec A – cos A ) sec2A = tan A. [4] [2011]

Step-by-step Explanation:

$LHS\\=(\cos ec\;A\;–\;\sin\;A)\;(sec\;A\;–\;\cos\;A\;)\;sec^2A\\=\left(\frac1{\sin A}-\sin A\right)\;\left(\frac1{\cos A}-\cos A\right)\;\frac1{\cos^2A}\\=\left(\frac{1-\sin^2A}{\sin A}\right)\;\left(\frac{1-\cos^2A}{\cos A}\right)\;\frac1{\cos^2A}\\=\left(\frac{\cos^2A}{\sin A}\right)\;\left(\frac{\sin^2A}{\cos A}\right).\;\frac1{\cos^2A}\\=\frac{\sin A}{\cos A}\\=\;\tan A\\=\;RHS\\Proved.$

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