Proving Trigonometric Identities : Exercise (10.3) - Solutions


Basic Trigonometric Identities

$\begin{array}{|l|} \hline \sin A=\displaystyle \frac{1}{\csc A}\ \text{and}\\ \csc A=\displaystyle \frac{1}{\sin A}\\ \cos A=\displaystyle \frac{1}{\sec A}\ \text{and}\\ \sec A=\displaystyle \frac{1}{\cos A}\\ \tan A=\displaystyle \frac{\sin A}{\cos A}\\ \tan A=\displaystyle \frac{1}{\cot A}\ \text{and}\\ \cot A=\displaystyle \frac{1}{\tan A}\\ \cot A=\displaystyle \frac{\cos A}{\sin A}\\ \hline\end{array}$

Pythagorean Identities

$\begin{array}{|l|} \hline \sin^2 A+ \cos^2 A=1\\ \tan^2 A+ 1=\sec^2 A\\ 1+ \cot^2 A=\csc^2 A\\ \hline\end{array}$

Trigonometric Ratios of Complementary Angles

$\begin{array}{|l|}\hline \sin \left(90^{\circ}-\alpha\right)=\cos \alpha\\ \cos \left(90^{\circ}-\alpha\right)=\sin \alpha\\ \tan \left(90^{\circ}-\alpha\right)=\cot \alpha\\ \cot \left(90^{\circ}-\alpha\right)=\tan \alpha\\ \sec \left(90^{\circ}-\alpha\right)=\csc \alpha\\ \csc \left(90^{\circ}-\alpha\right)=\sec \alpha\\ \hline \end{array}$



Prove the following identities.

1.           $\cot \theta \sqrt{1-\cos ^{2} \theta}=\cos \theta$

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$$\begin{aligned} &\ \ \ \ \ \cot \theta \sqrt{1-\cos ^{2} \theta}\\\\ &=\cot \theta \sqrt{\sin ^{2} \theta}\\\\ &=\frac{\cos \theta}{\sin \theta} \times \sin \theta\\\\ &=\cos \theta\end{aligned}$$

2.           $\displaystyle \frac{\tan ^{2} \theta+1}{\tan \theta \csc ^{2} \theta}=\tan \theta$

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$$\begin{aligned} &\ \ \ \ \ \frac{\tan ^{2} \theta+1}{\tan \theta \csc ^{2} \theta}\\\\ &=\frac{\sec ^{2} \theta}{\cos \theta} \times \frac{1}{\sin ^{2} \theta}\\\\ &=\frac{1}{\cos ^{2} \theta} \times \sin \theta \cos \theta\\\\ &=\frac{\sin \theta}{\cos \theta}=\tan \theta \end{aligned}$$

3.           $\left(1-\sin ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)=\cot ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \left(1-\sin ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)\\\\ &=\cos ^{2} \theta \csc ^{2} \theta\\\\ &=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\\\\ &=\cot ^{2} \theta\end{aligned}$

4.           $\tan ^{2} \theta-\cot ^{2} \theta=\sec ^{2} \theta-\csc ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \tan ^{2} \theta-\cot ^{2} \theta\\\\ & =\left(\sec ^{2} \theta-1\right)-\left(\csc ^{2} \theta-1\right)\\\\ &=\sec ^{2} \theta-1-\csc ^{2} \theta+1\\\\ &=\sec ^{2} \theta-\csc ^{2} \theta\end{aligned}$

5.           $\sin \theta \sec \theta \sqrt{\csc ^{2} \theta-1}=1$

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$\begin{aligned} &\ \ \ \ \ \sin \theta \sec \theta \sqrt{\csc ^{2} \theta-1}\\\\ &=\frac{\sin \theta}{\cos \theta} \times \sqrt{\cot ^{2} \theta}\\\\ &=\tan \theta \cot \theta=1\end{aligned}$

6.           $16 \sec ^{2} \theta+\csc ^{2} \theta=\sec ^{2} \theta \csc ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \sec ^{2} \theta+\csc ^{2} \theta \\\\ &=\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\\\\ &=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta \sin ^{2} \theta}\\\\ &=\frac{1}{\cos ^{2} \theta \sin ^{2} \theta}\\\\ &=\sec ^{2} \theta \csc ^{2} \theta\end{aligned}$

7.           $\left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)=1$

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$\begin{aligned} &\ \ \ \ \ \left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)\\\\ &=\sec ^{2} \theta \cos ^{2} \theta\\\\ &=\frac{1}{\cos ^{2} \theta} \times \cos ^{2} \theta\\\\ &=1\end{aligned}$

8.           $(1+\tan \theta)^{2}+(1-\tan \theta)^{2}=2 \sec ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ (1+\tan \theta)^{2}+(1-\tan \theta)^{2}\\\\ &=\left(1+2 \tan \theta+\tan ^{2} \theta\right)+\left(1-2 \tan \theta+\tan ^{2} \theta\right)\\\\ &=2+2 \tan ^{2} \theta\\\\ &=2\left(1+\tan ^{2} \theta\right)\\\\ &=2 \sec ^{2} \theta\end{aligned}$

9.           $\sec ^{2} \theta \cot ^{2} \theta-1=\cot ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \sec ^{2} \theta \cot ^{2} \theta-1 \\\\ &=\frac{1}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1\\\\ &=\frac{1}{\sin ^{2} \theta}-1\\\\ &=\csc ^{2} \theta-1\\\\ &=\cot ^{2} \theta\end{aligned}$

10.           $\displaystyle \frac{1}{1-\sin \theta}+\displaystyle \frac{1}{1+\sin \theta}=2 \sec ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\\\\ &=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}\\\\ &=\frac{2}{1-\sin ^{2} \theta}\\\\ &=\frac{2}{\cos ^{2} \theta}\\\\ &=2 \sec ^{2} \theta\end{aligned}$

11.           $\displaystyle \frac{1}{\sin ^{2} \theta}-\displaystyle \frac{1}{\tan ^{2} \theta}=1$

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$\begin{aligned} &\ \ \ \ \ \frac{1}{\sin ^{2} \theta}-\frac{1}{\tan ^{2} \theta}\\\\ &=\csc ^{2} \theta-\cot ^{2} \theta=1\end{aligned}$

12.           $(\tan \theta+\sec \theta)^{2}=\displaystyle \frac{1+\sin \theta}{1-\sin \theta}$

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$\begin{aligned} &\ \ \ \ \ (\tan \theta+\sec \theta)^{2}\\\\ &=\left(\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)^{2}\\\\ &=\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}\\\\ &=\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}\\\\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}\\\\ &=\frac{(1+\sin \theta)^{2}}{(1-\sin \theta)(1+\sin \theta)}\\\\ &=\frac{1+\sin \theta}{1-\sin \theta}\end{aligned}$

13.           $\sin ^{4} \theta-\cos ^{4} \theta=1-2 \cos ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \sin ^{4} \theta-\cos ^{4} \theta \\\\ &=\left(\sin ^{2} \theta\right)^{2}-\left(\cos ^{2} \theta\right)^{2}\\\\ &=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta-\cos ^{2} \theta\right) \\\\ &=1\left(1-\cos ^{2} \theta-\cos ^{2} \theta\right)\\\\ &=1-2 \cos ^{2} \theta\end{aligned}$

14.           $\displaystyle \frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}=\csc ^{2} \theta$

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$\begin{aligned} &\ \ \ \ \ \frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}\\\\ &=\frac{\tan ^{2} \theta}{\tan ^{2} \theta}+\frac{1}{\tan ^{2} \theta}\\\\ &=1+\cot ^{2} \theta\\\\ &=\csc ^{2} \theta\end{aligned}$

15.           $\sin ^{2} \theta \tan \theta+\cos ^{2} \theta \cot \theta+2 \sin \theta \cos \theta=\tan \theta+\cot \theta$

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$\begin{aligned} &\ \ \ \ \ {{\sin }^{2}}\theta \tan \theta +{{\cos }^{2}}\theta \cot \theta +2\sin \theta \cos \theta \\ &={{\sin }^{2}}\theta \cdot \frac{{\sin \theta }}{{\cos \theta }}+{{\cos }^{2}}\theta \cdot \frac{{\cos \theta }}{{\sin \theta }}+2\sin \theta \cos \theta \\ &=\frac{{{{{\sin }}^{3}}\theta }}{{\cos \theta }}+\frac{{{{{\cos }}^{3}}\theta }}{{\sin \theta }}+2\sin \theta \cos \theta \\ &=\frac{{{{{\sin }}^{4}}\theta +{{{\cos }}^{4}}\theta +2{{{\sin }}^{2}}\theta {{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)}}^{2}}}}{{\sin \theta \cos \theta }}\\ &=\frac{{\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)}}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\sin }}^{2}}\theta }}{{\sin \theta \cos \theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\tan \theta +\cot \theta \end{aligned}$

16.           Find the value of acute angle $\alpha$ in each of the following equations:

               (a)    $\cos 2 \alpha=\sin 7 \alpha$

               (b)    $\tan 3 \alpha=\cot 2 \alpha$

               (c)    $\sec \alpha=\csc 5 \alpha$

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$\begin{array}{l} \text {(a)} \quad \cos 2 \alpha=\sin 7 \alpha\\\\ \therefore \quad 2\alpha + 7 \alpha = 90^{\circ}\\\\ \quad\quad 9\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 90^{\circ}\\\\ \text {(b)} \quad \tan 3 \alpha=\cot 2 \alpha\\\\ \therefore \quad 3\alpha + 2 \alpha = 90^{\circ}\\\\ \quad\quad 5\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 18^{\circ}\\\\ \text {(c)} \quad \sec \alpha=\csc 5 \alpha\\\\ \therefore \quad \alpha + 5 \alpha = 90^{\circ}\\\\ \quad\quad 6\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 15^{\circ} \end{array}$

17.           Prove the identity $\cos \left(90^{\circ}-\alpha\right) \tan \left(90^{\circ}-\alpha\right)=\cos \alpha$

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$\begin{aligned} &\ \ \ \ \ \cos (90-\alpha) \tan (90-\alpha)\\\\ &=\sin \alpha \cot \alpha\\\\ &=\sin \alpha \times \frac{\cos \alpha}{\sin \alpha}\\\\ &=\cos \alpha \end{aligned}$

18.           Prove the identity $\sin \left(90^{\circ}-\alpha\right) \sec \left(90^{\circ}-\alpha\right)=\cot \alpha$

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$\begin{aligned} &\ \ \ \ \ \sin (90-\alpha) \sec \left(90^{\circ}-\alpha\right)\\\\ &=\cos \alpha \csc \alpha\\\\ &=\cos \alpha \times \frac{1}{\sin \alpha}\\\\ &=\frac{\cos \alpha}{\sin \alpha}\\\\ &=\cot \alpha\end{aligned}$