Class 10 Trigonometrical Identities Mock Test (ICSE)

Topics in Trigonometrical Identities

• Fundamental identities
• Proving identities
• Trigonometric ratios of complementary angles
• Simplifying trigonometric expressions

Trigonometrical Identities practice questions with answers

1. Q1Multiple choice·1 mark
   What is the value of $(1 - \sin^2\theta) \sec^2\theta$?

   1. (a)$\sin^2\theta$
   2. (b)$0$
   3. (c)$1$
   4. (d)$-1$
   Show answer and step marks

   1. Correct option: (c) $1$. Using the fundamental identity $1 - \sin^2\theta = \cos^2\theta$, the expression becomes $\cos^2\theta \cdot \sec^2\theta = \cos^2\theta \cdot \frac{1}{\cos^2\theta} = 1$.+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   If $\tan\theta + \sin\theta = m$ and $\tan\theta - \sin\theta = n$, which of the following relations is true?

   1. (a)$m^2 - n^2 = 2mn$
   2. (b)$m^2 - n^2 = 4\sqrt{mn}$
   3. (c)$m^2 - n^2 = 4mn$
   4. (d)$m^2 - n^2 = mn$
   Show answer and step marks

   1. Correct option: (b) $4\sqrt{mn}$. Computing $m^2 - n^2 = (\tan\theta + \sin\theta)^2 - (\tan\theta - \sin\theta)^2 = 4\tan\theta\sin\theta$, and $4\sqrt{mn} = 4\sqrt{\tan^2\theta - \sin^2\theta} = 4\sqrt{\sin^2\theta(\frac{1}{\cos^2\theta} - 1)} = 4\sin\theta\sqrt{\tan^2\theta} = 4\tan\theta\sin\theta$.+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   The value of $\frac{\tan 26^\circ}{\cot 64^\circ} + \frac{\cos 35^\circ}{\sin 55^\circ}$ is:

   1. (a)$0$
   2. (b)$\frac{1}{2}$
   3. (c)$2$
   4. (d)$1$
   Show answer and step marks

   1. Correct option: (c) $2$. Since $\cot 64^\circ = \tan(90^\circ - 64^\circ) = \tan 26^\circ$ and $\sin 55^\circ = \cos(90^\circ - 55^\circ) = \cos 35^\circ$, the expression simplifies to $\frac{\tan 26^\circ}{\tan 26^\circ} + \frac{\cos 35^\circ}{\cos 35^\circ} = 1 + 1 = 2$.+1

   Answer: (c)
4. Q4Multiple choice·1 mark
   Simplify the expression: $(\sec A + \tan A)(1 - \sin A)$.

   1. (a)$\sec A$
   2. (b)$\cos A$
   3. (c)$\sin A$
   4. (d)$\tan A$
   Show answer and step marks

   1. Correct option: (b) $\cos A$. Rewriting in sine and cosine gives $(\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A) = \frac{1 + \sin A}{\cos A}(1 - \sin A) = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$.+1

   Answer: (b)
5. Q5Long answer·4 marks
   Prove the following trigonometric identity for $0^\circ < \theta < 90^\circ$, $\theta \neq 45^\circ$: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$
   Show answer and step marks

   1. Express $\tan \theta$ as $\frac{\sin \theta}{\cos \theta}$ and $\cot \theta$ as $\frac{\cos \theta}{\sin \theta}$ on the L.H.S.+1
   2. Rewrite the expression as $\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$ and simplify the denominators: $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}$.+1
   3. Take the LCM and combine the terms: $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$ and apply the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.+1
   4. Cancel $(\sin \theta - \cos \theta)$ from numerator and denominator to get $\frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \csc \theta \sec \theta + 1 = 1 + \sec \theta \csc \theta$, hence proved.+1

   Answer: Hence proved

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