Class 10: Trigonometry – Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: } \frac{\sec A -1 }{\sec A + 1} = \frac{1- \cos A }{1+ \cos A} \hspace{1.0cm} [2007]$

Answer:

$\displaystyle \text{LHS } =$ $\displaystyle \frac{\sec A -1 }{\sec A + 1}$

$\displaystyle =$ $\displaystyle \frac{\frac{1}{\cos A} -1 }{\frac{1}{\cos A} + 1}$

$\displaystyle =$ $\displaystyle \frac{1- \cos A }{1+ \cos A} =$ RHS. Hence Proved.

$\displaystyle \\$

$\displaystyle \text{Question 2: } (1-\tan A)^2 + (1 + \tan A)^2 = 2 \sec^2 A \hspace{1.0cm} [2005]$

Answer:

$\displaystyle \text{LHS } =$ $\displaystyle (1-\tan A)^2 + (1 + \tan A)^2$

$\displaystyle =$ $\displaystyle (1- \frac{\sin A}{\cos A})^2 + (1+\frac{\sin A}{\cos A})^2$

$\displaystyle =$ $\displaystyle \frac{(\cos A - \sin A)^2}{\cos^2 A} + \frac{(\cos A + \sin A)^2}{\cos^2 A}$

$\displaystyle =$ $\displaystyle \frac{\cos^2 A + \sin^2 A - 2 \cos A . \sin A+ \cos^2 A + \sin^2 A + 2 \cos A . \sin A}{\cos^2 A}$

$\displaystyle =$ $\displaystyle \frac{2}{\cos^2 A} = 2 \sec^2 A =$ RHS. Hence Proved.

$\displaystyle \\$

$\displaystyle \text{Question 3: } \frac{\sin A}{1 + \cos A} = \mathrm{cosec} A - \cot A \hspace{1.0cm} [2008]$

Answer:

$\displaystyle \text{RHS } = \mathrm{cosec} A - \cot A$

$\displaystyle = \frac{1}{\sin A} - \frac{\cos A}{\sin A}$

$\displaystyle = \frac{1- \cos A}{\sin A}$

$\displaystyle = \frac{1- \cos A}{\sin A} \times \frac{1+ \cos A}{1+ \cos A}$

$\displaystyle = \frac{1 - \cos^2 A}{\sin A (1 + \cos A)}$

$\displaystyle = \frac{\sin A}{1 + \cos A} =$ LHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 4: } \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \mathrm{cosec} A - \cot A \hspace{1.0cm} [2000]$

Answer:

$\displaystyle \text{LHS } = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$\displaystyle = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}}$

$\displaystyle = \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}}$

$\displaystyle = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}}$

$\displaystyle = \frac{1- \cos A}{\sin A}$

$\displaystyle = \mathrm{cosec} A - \cot A =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 5: } \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \hspace{1.0cm} [2013]$

Answer:

$\displaystyle \text{LHS } = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$

$\displaystyle = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 + \cos A}{1 + \cos A}}$

$\displaystyle = \sqrt{\frac{1 - \cos^2 A}{(1 + \cos A)^2}}$

$\displaystyle = \frac{\sin A}{1 + \cos A}$ = RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 6: } 1 - \frac{\cos^2 A}{1 + \sin A} = \sin A \hspace{1.0cm} [2001]$

Answer:

$\displaystyle \text{LHS } = \frac{\cos^2 A}{1 + \sin A}$

$\displaystyle = \frac{1 + \sin A - \cos^2 A}{1 + \sin A}$

$\displaystyle = \frac{\sin A + \sin^2 A}{1 + \sin A}$

$\displaystyle = \frac{\sin A(1 + \sin A)}{1 + \sin A}$

$\displaystyle = \sin A =$ RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 7: } \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A} = \frac{2 \sin A}{1 - 2 \cos^2 A} \hspace{1.0cm} [2002]$

Answer:

$\displaystyle \text{LHS } = \frac{1}{\sin A + \cos A} + \frac{1}{\sin A - \cos A}$

$\displaystyle = \frac{\sin A - \cos A + \sin A + \cos A}{\sin^2 A - \cos^2 A}$

$\displaystyle = \frac{\sin A - \cos A + \sin A + \cos A}{1 - \cos^2 A - \cos^2 A}$

$\displaystyle = \frac{2 \sin A}{1 - 2 \cos^2 A}$ = RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 8: } \frac{\sin \theta . \tan \theta}{1 - \cos \theta} = 1 + \sec \theta \hspace{1.0cm} [2006]$

Answer:

$\displaystyle \text{LHS } = \frac{\sin \theta . \tan \theta}{1 - \cos \theta}$

$\displaystyle = \frac{\sin^2 \theta}{\cos \theta (1 - \cos \theta)}$

$\displaystyle = \frac{(1-\cos \theta)(1 + \cos \theta)}{\cos \theta (1 - \cos \theta)}$

$\displaystyle = \frac{1 + \cos \theta}{\cos \theta}$

$\displaystyle = \sec \theta + 1$ = RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 9: } \frac{2 \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o} \hspace{1.0cm} [2006]$

Answer:

$\displaystyle \frac{2 \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o}$

$\displaystyle = \frac{2 \tan (90^o - 37^o)}{\cot 37^o}-\frac{\cot (90^o - 10^o)}{\tan 10^o}$

$\displaystyle = \frac{2 \cot 37^o}{\cot 37^o}-\frac{\tan 10^o}{\tan 10^o} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \cos^2 26^o + \cos 64^o.\sin 26^o + \frac{\tan 36^o}{\cot 54^o} \hspace{1.0cm} [2012]$

Answer:

$\displaystyle \cos^2 26^o + \cos 64^o.\sin 26^o + \frac{\tan 36^o}{\cot 54^o}$

$\displaystyle = \cos^2 26^o + \cos (90^o - 26^o).\sin 26^o + \frac{\tan (90^o - 54^o)}{\cot 54^o}$

$\displaystyle = \cos^2 26^o + \sin^2 26^o + \frac{\cot 54^o}{\cot 54^o}$

$\displaystyle = 2$

$\displaystyle \\$

$\displaystyle \text{Question 11: } 3 \cos 80^o . \mathrm{cosec} 10^o + 2 \sin 59^o.\sec 31^o \hspace{1.0cm} [2013]$

Answer:

$\displaystyle 3 \cos 80^o . \mathrm{cosec} 10^o + 2 \sin 59^o.\sec 31^o$

$\displaystyle = 3 \cos 80^o . \mathrm{cosec} (90^o - 80^o) + 2 \sin 59^o.\sec (90^o - 59^o)$

$\displaystyle = 3 \cos 80^o . \sec 80^o + 2 \sin 59^o . \mathrm{cosec} 59^o$

$\displaystyle = 3 + 2 = 5$

$\displaystyle \\$

$\displaystyle \text{Question 12: } \frac{\sin 80^o}{\cos 10^o} + \sin 59^o . \sec 31^o \hspace{1.0cm} [2007]$

Answer:

$\displaystyle \frac{\sin 80^o}{\cos 10^o} + \sin 59^o . \sec 31^o$

$\displaystyle = \frac{\sin (90^o - 10^o)}{\cos 10^o} + \sin 59^o . \sec (90^o - 59^o)$

$\displaystyle = \frac{\cos 10^o}{\cos 10^o} + \sin 59^o . \mathrm{\mathrm{cosec} } 59^o = 2$

$\displaystyle \\$

$\displaystyle \text{Question 13: } 14 \sin 30^o + 6 \cos 60^o - 5 \tan 45^o \hspace{1.0cm} [2004]$

Answer:

$\displaystyle 14 \sin 30^o + 6 \cos 60^o - 5 \tan 45^o$

$\displaystyle = 14 \sin (90^o - 60^o) + 6 \cos 60^o - 5 \tan 45^o$

$\displaystyle = 14 \cos 60^o + 6 \cos 60^o - 5 \tan 45^o$

$\displaystyle = 20 \cos 60^o - 5 \tan 45^o$

$\displaystyle = 20 \times \frac{1}{2} - 5 \times 1$

$\displaystyle = 10-5 = 5$

$\displaystyle \\$

Question 14: Evaluate without using trigonometric tables:

$\displaystyle 2 (\frac{\tan 35^o}{\cot 55^o})^2 + (\frac{\cot 55^o}{\tan 35^o})^2 - 3 (\frac{\sec 40^o}{\mathrm{cosec} 50^o}) \hspace{1.0cm} [2011]$

Answer:

$\displaystyle 2 (\frac{\tan 35^o}{\cot 55^o})^2 + (\frac{\cot 55^o}{\tan 35^o})^2 - 3 (\frac{\sec 40^o}{\mathrm{cosec} 50^o})$

$\displaystyle = 2 (\frac{\tan 35^o}{\cot (90^o - 35^o)})^2 + (\frac{\cot 55^o}{\tan (90^o - 55^o)})^2 - 3 (\frac{sec 40^o}{\mathrm{cosec} (90^o - 40^o)})$

$\displaystyle = 2 (\frac{\tan 35^o}{\tan 35^o})^2 + (\frac{\cot 55^o}{\cot 55^o})^2 - 3 (\frac{\sec 40^o}{\sec 40^o})$

$\displaystyle = 2 + 1 - 3 = 0$

$\displaystyle \\$

$\displaystyle \text{Question 15: Prove that } (\mathrm{cosec} A- \sin A) (\sec A-\cos A) \sec^2 A = \tan A \hspace{1.0cm} [2011]$

Answer:

$\displaystyle (\mathrm{cosec} A- \sin A) (\sec A-\cos A) \sec^2 A = \tan A$

$\displaystyle \text{LHS } = (\mathrm{cosec} A- \sin A) (\sec A-\cos A) \sec^2 A$

$\displaystyle = 2 \frac{1 - \sin^2 A}{\sin A} . \frac{1 - \cos^2 A}{\cos A} . \frac{1}{\cos^2 A}$

$\displaystyle = 2 \frac{\cos^2 A}{\sin A}. \frac{\sin^2 A}{\cos A} . \frac{1}{\cos^2 A}$

$\displaystyle = 2 \frac{\sin A}{\cos A} =\tan A =$ RHS

Hence Proved.

$\displaystyle \\$

Question 16: Without using trigonometric tables, evaluate

$\displaystyle \sin^2 34^o + \sin^2 56^o + 2 \tan 18^o \tan 72^o - \cot ^2 30^o \hspace{1.0cm} [2014]$

Answer:

$\displaystyle \text{Given } \sin^2 34^o + \sin^2 56^o + 2 \tan 18^o \tan 72^o - \cot ^2 30^o$

$\displaystyle = \sin^2 34^o + \sin^2 (90^o - 34^o) + 2 \tan 18^o \tan (90^o-18^o) - \cot ^2 30^o$

$\displaystyle = \sin^2 34^o + \cos^2 34^o + 2 \tan 18^o \cot 18^o - (\sqrt{3})^2$

$\displaystyle = 1+ 2 \tan 18^o \times \frac{1}{\tan 18^o}-3$

$\displaystyle = 1+2-3 = 0$

$\displaystyle \\$

Question 17: Prove the identity:

$\displaystyle (\sin \theta + \cos \theta) (\tan \theta + \cot \theta) = \sec \theta + \mathrm{cosec} \theta \hspace{1.0cm} [2014]$

Answer:

$\displaystyle \text{LHS } = (\sin \theta + \cos \theta)(\tan \theta+ \cot \theta)$

$\displaystyle = (\sin \theta + \cos \theta) (\frac{\sin \theta}{\cos \theta} + \frac{cos \theta}{\sin \theta})$

$\displaystyle = (\sin \theta + \cos \theta) (\frac{\sin^2 \theta+ \cos^2 \theta}{\sin \theta . \cos \theta})$

$\displaystyle = (\sin \theta + \cos \theta) (\frac{1}{\sin \theta . \cos \theta})$

$\displaystyle = (\frac{\sin \theta}{\sin \theta . \cos \theta}) + (\frac{\cos \theta}{sin \theta . \cos \theta})$

$\displaystyle = \frac{1}{\cos \theta}+\frac{1}{\sin \theta}$

$\displaystyle = \sec \theta+\mathrm{cosec} \theta = RHS$

Hence Proved.

$\displaystyle \\$

Question 18: If $\displaystyle 2 \sin A - 1 = 0$ , show that $\displaystyle \sin 3A = 2 \sin A - 4 \sin^3 A \hspace{1.0cm} [2001]$

Answer:

$\displaystyle 2 \sin A - 1 = 0$

$\displaystyle \Rightarrow \sin A = \frac{1}{2} \Rightarrow A = 30^o$

$\displaystyle \text{Therefore to prove: } \sin 3A = 2 \sin A - 4 \sin^3 A$

$\displaystyle \text{LHS } = \sin 3A = \sin 90^o = 1$

$\displaystyle \text{RHS } = 2 \sin A - 4 \sin^3 A = 2 \sin 30^o - 4 \sin^3 30^o = \frac{3}{2} - \frac{1}{2} = 1$

Therefore LHS = RHS. Hence proved.

$\displaystyle \\$

$\text{Question 19: Evaluate: } \displaystyle \frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{\mathrm{cosec} 58^o} \hspace{1.0cm} [2000]$

Answer:

$\displaystyle \frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{\mathrm{cosec} 58^o}$

$\displaystyle = \frac{3 \sin 72^o}{\cos (90^o - 72^o)}- \frac{\sec 32^o}{\mathrm{cosec} (90^o - 32^o)}$

$\displaystyle = \frac{3 \sin 72^o}{\sin 72^o}- \frac{\sec 32^o}{\sec 32^o}$

$\displaystyle = 3 - 1 = 2$

$\displaystyle \\$

$\displaystyle \text{Question 20: Evaluate: } 3 \cos 80^o \mathrm{cosec} 10^o + 2 \cos 59^o \mathrm{cosec} 31^o \hspace{1.0cm} [2002]$

Answer:

$\displaystyle 3 \cos 80^o \mathrm{cosec} 10^o + 2 \cos 59^o \mathrm{cosec} 31^o$

$\displaystyle = 3 \cos 80^o \mathrm{cosec} (90^o - 80^o) + 2 \cos 59^o \mathrm{cosec} (90^o - 59^o)$

$\displaystyle = 3 \cos 80^o \sec 80^o + 2 \cos 59^o \sec 59^o$

$\displaystyle = 3 + 2 = 5$

$\displaystyle \\$

$\displaystyle \text{Question 21: } \frac{\cos 75^o}{\sin 15^o} + \frac{\sin 12^o}{\cos 78^o} - \frac{\cos 18^o}{\sin 72^o} \hspace{1.0cm} [2003]$

Answer:

$\displaystyle \frac{\cos 75^o}{\sin 15^o} + \frac{\sin 12^o}{\cos 78^o} - \frac{\cos 18^o}{\sin 72^o}$

$\displaystyle = \frac{\cos 75^o}{\sin (90^o - 75^o)} + \frac{\sin 12^o}{\cos (90^o - 12^o)} - \frac{\cos 18^o}{\sin (90^o - 18^o)}$

$\displaystyle = \frac{\cos 75^o}{\cos 75^o} + \frac{\sin 12^o}{\sin 12^o} - \frac{\cos 18^o}{\cos 18^o}$

$\displaystyle = 1+ 1 - 1 = 1$

$\displaystyle \\$

$\displaystyle \text{Question 22: Prove that: } \frac{\sin A}{1 + \cos A} + \frac{1 +\cos A}{\sin A} = 2 \mathrm{cosec} A \hspace{1.0cm} [2009]$

Answer:

$\displaystyle \text{LHS } =$ $\displaystyle \frac{\sin A}{1 + \cos A} + \frac{1 +\cos A}{\sin A}$

$\displaystyle =$ $\displaystyle \frac{\sin^2 A + (1 + \cos A)^2}{(1+ \cos A)\sin A}$

$\displaystyle =$ $\displaystyle \frac{\sin^2 A + 1 + \cos^2 A + 2 \cos A}{(1+ \cos A)\sin A}$

$\displaystyle =$ $\displaystyle \frac{2(1+\cos A)}{(1+ \cos A)\sin A}$

$\displaystyle =$ $\displaystyle \frac{2}{\sin A}$

$\displaystyle =$ $\displaystyle 2 \mathrm{cosec} A =$ RHS.

Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 23: Prove that: } \frac{\cos A \cot A}{1 - \sin A} = 1 + \mathrm{cosec} A \hspace{1.0cm} [2006]$

Answer:

$\displaystyle \text{LHS } = \frac{\cos A. \cot A}{1 - \sin A}$

$\displaystyle = \frac{\cos A . \frac{\cos A}{\sin A}}{1 - \sin A}$

$\displaystyle = \frac{\cos^2 A}{\sin A(1 - \sin A)}$

$\displaystyle = \frac{(1-\sin A)(1 + \sin A)}{\sin A(1 - \sin A)}$

$\displaystyle = \frac{1 + \sin A}{\sin A}$

$\displaystyle = \mathrm{cosec} A + 1 =$ RHS. Hence proved.

$\displaystyle \\$

Question 24: Without using trigonometric tables evaluate :

$\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o} \hspace{1.0cm} [2010]$

Answer:

Given: $\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o}$

$\displaystyle = \frac{\sin 35^o \cos (90^o- 35^o) + \cos 35^o \sin (90^o- 35^o)}{\mathrm{cosec} ^2 10^o - \tan^2 (90^o - 10^o)}$

$\displaystyle = \frac{\sin^2 35^o+\cos^2 35^o}{1+ \cot^2 10^o - \cot^2 10^o}$

$\displaystyle = \frac{1}{1}$

$\displaystyle = 1$

$\displaystyle \\$

Question 25: Without using trigonometric tables evaluate :

$\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o} \hspace{1.0cm} [2010]$

Answer:

Given: $\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o}$

$\displaystyle = \frac{\sin 35^o \cos (90^o- 35^o) + \cos 35^o \sin (90^o- 35^o)}{\mathrm{cosec} ^2 10^o - \tan^2 (90^o - 10^o)}$

$\displaystyle = \frac{\sin^2 35^o+\cos^2 35^o}{1+ \cot^2 10^o - \cot^2 10^o}$

$\displaystyle = \frac{1}{1} = 1$

11 thoughts on “Class 10: Trigonometry – Board Problems”

i2qtech/youtube says:

August 1, 2019 at 6:51 pm

The qustion no.2
Incorrect..

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1. ICSE CBSE Mathematics Student-Assistant says:
  
  August 2, 2019 at 8:32 pm
  
  Thank you for pointing out. It was a typing mistake. I have corrected it.
  
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Krishna says:

April 11, 2020 at 5:59 pm

It’s unable to understand the solution of question no.7

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1. ICSE CBSE ISC Board Mathematics Portal for Students says:
  
  April 11, 2020 at 7:56 pm
  
  There was a typing mistake in the question… i have added one more line of explanation…. it should be easy to understand now.
  
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2. ICSE CBSE ISC Board Mathematics Portal for Students says:
  
  July 19, 2020 at 9:31 am
  
  I fixed it. Thank you for your contribution.
  
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Vardaan says:

June 5, 2020 at 11:09 am

Technical issues from Q 12
Please sort them out

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1. ICSE CBSE ISC Board Mathematics Portal for Students says:
  
  June 5, 2020 at 12:13 pm
  
  Let me check
  
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2. ICSE CBSE ISC Board Mathematics Portal for Students says:
  
  July 19, 2020 at 9:29 am
  
  I fixed it. Thank you for your contribution.
  
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Palak says:

January 1, 2021 at 11:20 pm

Very helpful.

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suraj says:

January 9, 2021 at 1:22 pm

ques 9 has
wrong answer

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1. ICSE CBSE ISC Board Mathematics Portal for Students says:
  
  January 9, 2021 at 7:24 pm
  
  Good catch… it was a typo. We have corrected it.
  
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Class 10: Trigonometry – Board Problems

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