Note: This is the Solution of review exercise 6 from newly published book by PCTB (Punjab Curriculum and Textbook Board, Pakistan) for new 9th session 2025 Onward.
Review Exercise 6
Question # 01: Four option are given against each statement. Encircle the correct option.
\((i)\) The value of \(\tan^{-1} 2\) in radians is:
\( a) \quad \) \( \frac{\pi}{2} \)
\( b) \quad \) \( \frac{3\pi}{2}\)
\(c) \quad \) \( 0.463\pi\)
\( d) \quad \) \( 0.4636\)
\( \text{Answer/Explanation}\)
\((ii)\) In a right triangle, the hypotenuse is \(13 \) units and one of the angle is \(\theta=30^\circ\). The length of the opposite side is:
\( a) \quad \) \( 6.5\) units
\( b) \quad \) \( 7.5\) units
\(c) \quad \) \(6 \) units
\( d) \quad \) \(5 \) units
\( \text{Answer/Explanation}\)
\((iii)\) A person standing \(50\ m\) away from a building sees the top of the building at an angle of elevation of \(45^\circ\). Height of the building is:
\( a) \quad \) \( 50\) m
\( b) \quad \) \(25 \) m
\(c) \quad \) \(35 \) m
\( d) \quad \) \( 70\) m
\( \text{Answer/Explanation}\)
\((iv)\)\(\quad\)\(\sec^2 \theta -\tan^2\theta =\)_________.
\( a) \quad \) \(\sin^2 \theta\)
\( b) \quad \) \(1\)
\(c) \quad \) \(\cos^2 \theta\)
\( d) \quad \) \(\cot^2 \theta\)
\( \text{Answer/Explanation}\)
\((v)\) \(\quad\)If \(\sin \theta =\frac{3}{5}\) is an acute angle, \(\cos^2 \theta=\)____________.
\( a) \quad \) \(\frac{7}{25}\)
\( b) \quad \) \(\frac{24}{25}\)
\(c) \quad \) \(\frac{16}{25}\)
\( d) \quad \) \(\frac{4}{25}\)
\( \text{Answer/Explanation}\)
\((vi)\) \(\frac{5\pi}{24}\) rad \(=\) __________ degrees.
\( a) \quad \) \(30^\circ\)
\( b) \quad \) \(37.5^\circ\)
\(c) \quad \) \(45^\circ\)
\( d) \quad \) \(52.5^\circ\)
\( \text{Answer/Explanation}\)
\((vii)\)\(\quad\)\(292.5^\circ=\)___________ rad.
\( a) \quad \) \(\frac{17\pi}{6}\)
\( b) \quad \) \(\frac{17\pi}{4}\)
\(c) \quad \) \(1.6\pi\)
\( d) \quad \) \(1.625\pi\)
\( \text{Answer/Explanation}\)
\((viii)\)\(\quad\)Which of the following is a valid identity?
\( a) \quad \) \( \cos(\frac{\pi}{2}-\theta)=\sin \theta\)
\( b) \quad \) \( \cos(\frac{\pi}{2}-\theta)=\cos \theta\)
\(c) \quad \) \( \cos(\frac{\pi}{2}-\theta)=\sec \theta\)
\( d) \quad \) \( \cos(\frac{\pi}{2}-\theta)=\csc \theta\)
\( \text{Answer/Explanation}\)
\((ix)\)\(\quad\)\(\sin 60^\circ=\)__________.
\( a) \quad \) \(1\)
\( b) \quad \) \(\frac{1}{2}\)
\(c) \quad \) \(\sqrt{(3)^2}\)
\( d) \quad \) \(\frac{\sqrt{3}}{2}\)
\( \text{Answer/Explanation}\)
\((x)\) \(\quad\)\(\cos^2 100\pi+\sin^2 100\pi=\)____________:
\( a) \quad \) \(1\)
\( b) \quad \) \(2\)
\(c) \quad \) \(3\)
\( d) \quad \) \(4\)
\( \text{Answer/Explanation}\)
Question 2: Convert the given angle from:
\(\small{(a)}\) \(\quad\) degrees to radians giving answer in terms of \(\small{\pi}\)
\((i)\) \(\quad\) \(255^\circ\)
\((ii)\) \(\quad\) \(75^\circ{45}’ \)
\((iii)\) \(\quad\) \(142.5^\circ\)
\(\small{(b)}\) \(\quad\) radians to degrees giving answer in degree and minutes.
\((i)\) \(\quad\) \(\frac{17\pi}{24}\)
\((ii)\) \(\quad\) \(\frac{7\pi}{12} \)
\((iii)\) \(\quad\) \(\frac{11\pi}{16}\)
Question 3: Prove the following trigonometric identities:
\((i)\) \(\quad\) \(\large{\frac{\sin \theta}{1-\cos \theta}}=\large{\frac{1+\cos \theta}{\sin \theta}}\)
\((ii)\) \(\quad\) \(\sin \theta (\csc \theta -\sin \theta)=\large{\frac{1}{\sec^2 \theta}}\)
\((iii)\) \(\quad\) \(\large{\frac{\csc \theta -\sec \theta}{\csc \theta +\sec \theta}}=\large{\frac{1-\cos \theta}{1+\cos \theta} }\)
\((iv)\) \(\quad\) \(\tan \theta+\cot \theta=\large{\frac{1}{\sin \theta \cos \theta }}\)
\((v)\) \(\quad\) \(\large{\frac{\cos \theta+\sin \theta}{\cos \theta -\sin \theta }}+\large{\frac{\cos \theta -\sin \theta}{\cos \theta +\sin \theta}}=\large{\frac{2}{1-2\sin^2 \theta }}\)
\((vi)\) \(\quad\) \(\large{\frac{1+\cos \theta}{1-\cos \theta }}=(\csc \theta +\cot \theta)^2\)