Class 9 Trigonometrical Ratios Mock Test (ICSE)

Topics in Trigonometrical Ratios

• Definitions of the six ratios
• Reciprocal relations
• Finding ratios from one given ratio
• Relations between ratios

Trigonometrical Ratios practice questions with answers

1. Q1Multiple choice·1 mark
   If $\theta$ is an acute angle and $\sin \theta = \frac{5}{13}$, what is the value of $\cos \theta$?

   1. (a)$\frac{12}{13}$
   2. (b)$\frac{5}{12}$
   3. (c)$\frac{12}{5}$
   4. (d)$\frac{13}{12}$
   Show answer and step marks

   1. Correct option: (a) $\frac{12}{13}$. As $\theta$ is acute, $\cos \theta = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   In a right-angled triangle $ABC$, right-angled at $B$, if $AB = 3$ cm and $BC = 4$ cm, what is the value of $\sin A$?

   1. (a)$\frac{3}{4}$
   2. (b)$\frac{4}{3}$
   3. (c)$\frac{4}{5}$
   4. (d)$\frac{3}{5}$
   Show answer and step marks

   1. Correct option: (c) $\frac{4}{5}$. By Pythagoras theorem, Hypotenuse $AC = \sqrt{3^2 + 4^2} = 5$ cm. Then $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4}{5}$.+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   What is the product of $\tan \theta$ and $\cot \theta$?

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

   1. Correct option: (c) $1$. Since $\cot \theta$ is the reciprocal of $\tan \theta$, their product is $\tan \theta \times \frac{1}{\tan \theta} = 1$.+1

   Answer: (c)
4. Q4Multiple choice·1 mark
   Which of the following expressions is identically equal to $1 - \sin^2 \theta$?

   1. (a)$\operatorname{cosec}^2 \theta$
   2. (b)$\sec^2 \theta$
   3. (c)$\tan^2 \theta$
   4. (d)$\cos^2 \theta$
   Show answer and step marks

   1. Correct option: (d) $\cos^2 \theta$. By the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we get $1 - \sin^2 \theta = \cos^2 \theta$.+1

   Answer: (d)
5. Q5Long answer·4 marks
   Prove that $\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta} = \left(\frac{1 - \tan \theta}{1 - \cot \theta}\right)^2 = \tan^2 \theta$, for any acute angle $\theta$ where $\theta \neq 0^{\circ}$, $\theta \neq 45^{\circ}$ and $\theta \neq 90^{\circ}$.
   Show answer and step marks

   1. Consider the first part: $\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta}$. Using the identity relations $1 + \tan^2 \theta = \sec^2 \theta$ and $1 + \cot^2 \theta = \csc^2 \theta$, the expression becomes $\frac{\sec^2 \theta}{\csc^2 \theta}$.+1
   2. Express $\sec^2 \theta$ and $\csc^2 \theta$ in terms of sine and cosine: $\frac{\frac{1}{\cos^2 \theta}}{\frac{1}{\sin^2 \theta}} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$.+1
   3. Now consider the middle part: $\left(\frac{1 - \tan \theta}{1 - \cot \theta}\right)^2$. Substitute $\cot \theta = \frac{1}{\tan \theta}$ inside the bracket to get $\left(\frac{1 - \tan \theta}{1 - \frac{1}{\tan \theta}}\right)^2$.+1
   4. Simplify the denominator: $\frac{1 - \tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} = -\tan \theta$. Squaring this yields $(-\tan \theta)^2 = \tan^2 \theta$. Hence, both expressions are equal to $\tan^2 \theta$.+1

   Answer: Both $\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta}$ and $\left(\frac{1 - \tan \theta}{1 - \cot \theta}\right)^2$ simplify to $\tan^2 \theta$.

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