Question 7, Exercise 10.2

Solutions of Question 7 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove the identity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$.

$$\begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\\ &=\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\left( 1 \right)\\ &=\cos 2\theta \quad \text{(By using double angle identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align}$$

Prove the identity $\tan \dfrac{\theta }{2}+co\operatorname{t}\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$.

$$\begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operatorname{t}\dfrac{\theta }{2}\\ &=\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}+\dfrac{\cos \dfrac{\theta }{2}}{\sin \dfrac{\theta }{2}}\\ &=\dfrac{{{\sin }^{2}}\dfrac{\theta }{2}+{{\sin }^{2}}\dfrac{\theta }{2}}{\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}\\ &=\dfrac{1}{\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \quad \because \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right)\\ &=\dfrac{2}{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}\\ &=\dfrac{2}{2\sin \theta }\quad (By \,using\, double\, angle\, identity)\end{align}$$

Prove the identity $\dfrac{1+\cos 2\theta }{1-\cos 2\theta }={{\cot }^{2}}\theta$.

$$\begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1+\cos 2\theta }\\ &=\dfrac{1+\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)}{1-\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)}\\ &=\dfrac{1+{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }\\ &=\dfrac{1-{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{1-{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }\\ &=\dfrac{2{{\cos }^{2}}\theta }{2si{{n}^{2}}\theta }={{\cot }^{2}}\theta =R.H.S.\end{align}$$

Prove the identity $\cos ec2\theta -\cot 2\theta =tan\theta$.

$$\begin{align}L.H.S.&=\cos ec2\theta -\cot 2\theta \\ &=\dfrac{1}{\sin 2\theta }-\dfrac{\cos 2\theta }{\sin 2\theta }\\ &=\dfrac{1-\cos 2\theta }{\sin 2\theta }\\ &=\dfrac{1-\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)}{\sin 2\theta }\\ &=\dfrac{1-{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\sin 2\theta }\\ &=\dfrac{2{{\sin }^{2}}\theta }{2\sin \theta \cos \theta }=\tan \theta =R.H.S.\end{align}$$

Prove the identity $\dfrac{\sin 3\beta }{\sin \beta }-\dfrac{\cos 3\beta }{\cos \beta }=2$.

$$\begin{align}L.H.S.&=\dfrac{\sin 3\beta }{\sin \beta }-\dfrac{\cos 3\beta }{\cos \beta }\\ &=\dfrac{\sin 3\beta \cos \beta -\cos 3\beta \sin \beta }{\sin \beta \cos \beta }\\ &=\dfrac{\sin \left( 3\beta -\beta \right)}{\sin \beta \cos \beta }\\ &=\dfrac{\sin 2\beta }{\sin \beta \cos \beta }=\dfrac{2\sin 2\beta }{2\sin \beta \cos \beta }=2=R.H.S.\end{align}$$

Prove the identity $\dfrac{\sin 3\theta }{\cos \theta }+\dfrac{\cos 3\theta }{\sin \theta }=2\cot 2\theta$.

$$\begin{align}L.H.S.&=\dfrac{\sin 3\theta }{\cos \theta }+\dfrac{\cos 3\theta }{\sin \theta }\\ &=\dfrac{\sin 3\theta sin\theta +\cos 3\theta \cos \theta }{\sin \theta \cos \theta }\\ &=\dfrac{\cos \left( 3\theta -\theta \right)}{\sin \theta \cos \theta }\\ &=\dfrac{\cos 2\theta }{\sin \theta \cos \theta }=\dfrac{2\cos 2\theta }{2\sin \theta \cos \theta }=2\cot 2\theta =R.H.S.\end{align}$$

Prove the identity $\dfrac{{{\cos }^{3}}\theta -{{\sin }^{3}}\theta }{\cos \theta -\sin \theta }=\dfrac{2+\sin 2\theta }{2}$.

$$\begin{align}L.H.S.&=\dfrac{{{\cos }^{3}}\theta -{{\sin }^{3}}\theta }{\cos \theta -\sin \theta }\\ &=\dfrac{\left( \cos \theta -\sin \theta \right)\left( {{\cos }^{2}}\theta +si{{n}^{2}}\theta +sin\theta \cos \theta \right)}{\cos \theta -\sin \theta }\\ &=1+sin\theta \cos \theta \\ &=\dfrac{2\left( 1+sin\theta \cos \theta \right)}{2}=\dfrac{2+sin2\theta }{2}=R.H.S.\end{align}$$

Prove the identity $\dfrac{2{{\cos }^{3}}\theta }{1-\sin \theta }=2\cos \theta +\sin 2\theta$.

$$\begin{align}L.H.S.&=\dfrac{2{{\cos }^{3}}\theta }{1-\sin \theta }\\ &=\dfrac{2\cos \theta \left( 1-{{\sin }^{2}}\theta \right)}{1-\sin \theta }\\ &=\dfrac{2\cos \theta \left( 1-\sin \theta \right)\left( 1+\sin \theta \right)}{1-\sin \theta }\\ &=2\cos \theta +2\cos \theta \sin \theta \\ &=2\cos \theta +2\sin 2\theta =R.H.S.\end{align}$$

Prove the identity $\cot 2\theta =\dfrac{1}{2}\left( \cot \theta -\dfrac{1}{\cot \theta } \right)$.

$$\begin{align}L.H.S.&=\cot 2\theta \\ &=\dfrac{\cos 2\theta }{\sin 2\theta }\\ &=\dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{2\sin \theta \cos \theta }\\ &=\dfrac{1}{2}\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta \cos \theta }-\dfrac{{{\sin }^{2}}\theta }{\sin \theta \cos \theta } \right)\\ &=\dfrac{1}{2}\left( \cot \theta -\tan \theta \right)\\ &=\dfrac{1}{2}\left( \cot \theta -\dfrac{1}{co\operatorname{t}\theta } \right)=R.H.S.\end{align}$$

Prove the identity $\dfrac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }+\dfrac{\sin \alpha -\cos \alpha }{\cos \alpha +\sin \alpha }=2\tan 2\alpha$.

$$\begin{align}L.H.S.&=\dfrac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }+\dfrac{\sin \alpha -\cos \alpha }{\cos \alpha +\sin \alpha }\\ &=\dfrac{{{\left( \sin \alpha +\cos \alpha \right)}^{2}}-{{\left( \sin \alpha -\cos \alpha \right)}^{2}}}{\left( \cos \alpha -\sin \alpha \right)\left( \cos \alpha +\sin \alpha \right)}\\ &=\dfrac{\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +2\sin \alpha \cos \alpha \right)-\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha -2\sin \alpha \cos \alpha \right)}{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha }\\ &=\dfrac{4\sin \alpha \cos \alpha }{\cos 2\alpha }=2\tan 2\alpha =R.H.S.\end{align}$$

Prove the identity $\tan \dfrac{\alpha }{2}=\dfrac{\sin \alpha }{1+\cos \alpha }$.

$$\begin{align}R.H.S.&=\dfrac{\sin \alpha }{1+\cos \alpha }\\ &=\dfrac{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}{2{{\cos }^{2}}\dfrac{\alpha }{2}}\quad(By\, using\, half\, angle\, identity\, and\, double\, angle\, identity)\\ &=\dfrac{\sin \dfrac{\alpha }{2}}{\cos \dfrac{\alpha }{2}}=\tan \dfrac{\alpha }{2}=R.H.S.\end{align}$$

Prove the identity $\dfrac{\cos ec\theta -\cot \theta }{1+\cos \theta }=\cos ec\theta {{\tan }^{2}}\dfrac{\theta }{2}$.

$$\begin{align}L.H.S.&=\dfrac{\cos ec\theta -\cot \theta }{1+\cos \theta }\\ &=\dfrac{\dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta }}{1+\cos \theta }\\ &=\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{\sin \theta 2{{\cos }^{2}}\dfrac{\theta }{2}}\\ &=\cos ec\theta {{\tan }^{2}}\dfrac{\theta }{2}=R.H.S.\end{align}$$

Prove the identity ${{\cos }^{2}}\dfrac{\theta }{2}=\dfrac{1-{{\cos }^{2}}\theta }{2-2\cos \theta }$.

$$\begin{align}R.H.S.&=\dfrac{1-{{\cos }^{2}}\theta }{2-2\cos \theta }\\ &=\dfrac{\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)}{2\left( 1-\cos \theta \right)}\\ &=\dfrac{\left( 1+\cos \theta \right)}{2}\\ &=\dfrac{2{{\cos }^{2}}\dfrac{\theta }{2}}{2}={{\cos }^{2}}\dfrac{\theta }{2}=L.H.S.\end{align}$$

Prove the identity $\dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}}=\cos \alpha$.

$$\begin{align}L.H.S.&=\dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}}\\ &=\dfrac{1-\dfrac{{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}}}{1+\dfrac{{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}}}\\ &=\dfrac{{{\cos }^{2}}\dfrac{\alpha }{2}-{{\sin }^{2}}\dfrac{\alpha }{2}}{{{\cos }^{2}}\dfrac{\alpha }{2}+{{\sin }^{2}}\dfrac{\alpha }{2}}\\ &=\cos \alpha =R.H.S.\end{align}$$

Prove the identity $\sin 2\theta -4{{\sin }^{3}}\theta \cos \theta =\sin 2\theta \cos 2\theta$.

$$\begin{align}L.H.S.&=\sin 2\theta -4{{\sin }^{3}}\theta \cos \theta \\ &=2\sin \theta \cos \theta -4{{\sin }^{3}}\theta \cos \theta \\ &=2\sin \theta \cos \theta \left( 1-2{{\sin }^{2}}\theta \right)\\ &=\sin 2\theta \cos 2\alpha =R.H.S.\end{align}$$

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