Note: This is the Solution of review exercise 6.3 from newly published book by PCTB (Punjab Curriculum and Textbook Board, Pakistan) for new 9th session 2025 Onward.
Exercise 6.3
Question # 01: If \(\small{\theta}\) lies in first quadrant, find the remaining trigonometric ratios of \(\small{\theta}\).
\((i)\) \(\quad\) \( \sin \theta =\frac{2}{3}\)
\((ii)\) \(\quad\) \(\cos \theta =\frac{3}{4}\)
\((iii)\) \(\quad\) \(\tan \theta =\frac{1}{2} \)
\((iv)\) \(\quad \)\(\sec \theta =3 \)
\((v)\) \(\quad \) \(\cot \theta =\sqrt{\frac{3}{2}} \)
Prove the following trigonometric identities:
Question 2: \(\small{(\sin \theta + \cos \theta)^2=1+2\sin\theta \cos \theta }\)
Question 3: \(\frac{\cos \theta}{\sin \theta }=\frac{1}{\tan \theta} \)
Question 4: \(\frac{\sin \theta}{\csc \theta }+\frac{\cos \theta}{\sec \theta}=1 \)
Question 5: \(\small{\cos^2 \theta\ -\ \sin^2 \theta =2\cos^2 \theta -1} \)
Question 6: \(\small{\cos^2 \theta\ -\ \sin^2 \theta =1-2\sin^2 \theta } \)
Question 7: \(\frac{1-\sin \theta}{\cos \theta}=\frac{\cos \theta}{1+\sin \theta} \)
Question 8: \(\small{(\sec \theta\ -\ \tan \theta)^2=}\frac{1-\sin \theta}{1+\sin \theta}\)
Question 9: \((\tan \theta +\cot \theta)^2=\sec^2 \theta \csc^2 \theta \)
Question 10: \(\frac{\tan \theta +\sec\theta-1}{\tan \theta -\sec \theta +1}=\small{\tan \theta +\sec \theta} \)
Question 11: \(\small{\sin^3 \theta -\cos^3 \theta =(\sin \theta -\cos \theta )(1+\sin\theta \cos \theta)}\)
Question 12: \(\small{\sin^6 \theta -\cos^6 \theta =(\sin^2 \theta -\cos^2 \theta)(1-\sin^2 \theta \cos^2 \theta)}\)
