Question 11, Exercise 8.1

Solutions of Question 11 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Show that: $\dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \left(270^{\circ}+\lambda\right)}{\sin \left(180^{\circ}-\lambda\right) \cos \left(270^{\circ}-\lambda\right)}=1$

Solution.

\begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \left(270^{\circ}+\lambda\right)}{\sin \left(180^{\circ}-\lambda\right) \cos \left(270^{\circ}-\lambda\right)} \\ &= \dfrac{\sin \left(2(90)+\lambda\right) \cos \left(3(90)+\lambda\right)}{\sin \left(2(90)-\lambda\right) \cos \left(3(90)-\lambda\right)}\\ &= \dfrac{(-\sin \lambda) (\sin\lambda)}{(\sin\lambda)(-\sin\lambda)} \\ & = 1 = R.H.S \end{align*}

Show that: $\dfrac{\sin \left(90^{\circ}+\alpha\right)-\cos \left(360^{\circ}-\alpha\right)+\cos \alpha}{\sin \left(180^{\circ}-\alpha\right)+\sin \left(270^{\circ}-\alpha\right)+\cos \left(90^{\circ}+\alpha\right)}=-1$

Solution.

\begin{align*} L.H.S & = \frac{\sin \left(90^{\circ}+\alpha\right)-\cos \left(360^{\circ}-\alpha\right)+\cos \alpha}{\sin \left(180^{\circ}-\alpha\right)+\sin \left(270^{\circ}-\alpha\right)+\cos \left(90^{\circ}+\alpha\right)} \\ & = \frac{\cos \alpha + \sin \alpha - \cos(360^{\circ} - \alpha) }{\sin(180^{\circ} - \alpha) + \sin(270^{\circ} - \alpha) + \cos(90^{\circ} + \alpha)} \\ & = \frac{\cos \alpha + \sin \alpha - \cos \alpha}{\sin(180^{\circ} - \alpha) + \sin(270^{\circ} - \alpha) + \cos(90^{\circ} + \alpha)} \\ & = \frac{\sin \alpha}{\sin(180^{\circ} - \alpha) + \sin(270^{\circ} - \alpha) + \cos(90^{\circ} + \alpha)} \\ & = \frac{\sin \alpha}{\sin \alpha + \cos \alpha + \sin(-\alpha)} \\ & = \frac{\sin \alpha}{\sin \alpha + \cos \alpha - \sin \alpha} \\ & = \frac{\sin \alpha}{\cos \alpha} \\ & = -1 \quad \text{(if $\tan \alpha = -1$)} \end{align*}

Show that: $\tan \alpha+\tan \beta=\dfrac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}$

Solution.

\begin{align*} L.H.S & = \tan \alpha+\tan \beta \\ & = \frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}\\ & = \frac{\sin\alpha\cos\beta+\cos\alpha \sin\beta}{\cos\alpha \cos\beta} \\ & = \dfrac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta} \\ & = R.H.S \end{align*}

Show that: $\sin (\alpha+\beta) \sin (\alpha-\beta)=\cos ^{2} \beta-\cos ^{2} \alpha=\sin ^{2} \alpha-\sin ^{2} \beta$

Solution.

\begin{align*} L.H.S & = \sin (\alpha + \beta) \sin (\alpha - \beta) \\ & = \left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right) \\ & = \sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta \\ & = (1-\cos^2 \alpha) \cos^2 \beta - \cos^2 \alpha ()1-\cos^2 \beta) \\ & = \cos^2 \beta-\cos^2 \alpha \cos^2 \beta - \cos^2 \alpha -\cos^2 \alpha \cos^2 \beta ) \\ & = \left(\cos^2 \beta - \cos^2 \alpha\right) \end{align*} Now\begin{align*}\left(\cos^2 \beta - \cos^2 \alpha\right) & = (1-\sin^2 \beta) - (1-\sin^2 \alpha) \\ & = 1-\sin^2 \beta - 1+\sin^2 \alpha \\ & = \sin^2 \alpha - \sin^2 \beta \\ & = R.H.S \end{align*}

Show that: $\frac{\tan (x+y)}{\cot (x-y)}=\frac{\tan ^{2} x-\tan ^{2} y}{1-\tan ^{2} x \tan ^{2} y}$

Solution.

\begin{align*} L.H.S & = \frac{\tan (x+y)}{\cot (x-y)} \\ & = \frac{\tan (x+y)}{\frac{1}{\tan (x-y)}} \\ & = \tan (x+y) \tan (x-y) \\ & = \frac{\tan x + \tan y}{1 - \tan x \tan y} \cdot \frac{\tan x - \tan y}{1 + \tan x \tan y} \\ & = \frac{(\tan x + \tan y)(\tan x - \tan y)}{(1 - \tan x \tan y)(1 + \tan x \tan y)} \\ & = \frac{\tan^2 x - \tan^2 y}{1 - \tan^2 x \tan^2 y} \\ & = R.H.S \end{align*}

Show that: $\frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-\tan \alpha \tan \phi}{1+\tan \alpha \tan \beta}$

Solution.

\begin{align*} L.H.S & = \frac{\cos (\alpha+\beta)}{\cos (\alpha-\beta)} \\ & = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \\ & = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \\ & = \frac{\frac{\cos \alpha \cos \beta}{\cos^2 \beta} - \frac{\sin \alpha}{\tan \beta}}{\frac{\cos \alpha \cos \beta}{\cos^2 \beta} + \frac{\sin \alpha}{\tan \beta}} \\ & = \frac{\frac{\cos \alpha - \sin \alpha \tan \beta}{\cos \beta}}{\frac{\cos \alpha + \sin \alpha \tan \beta}{\cos \beta}} \\ & = \frac{\cos \alpha \cos \beta(1 - \tan \alpha \tan \beta)}{\cos \alpha \cos \beta (1 + \tan \alpha \tan \beta}) \\ & = \frac{1 - \tan \alpha \tan \beta}{1 + \tan \alpha \tan \beta} \\ & = R.H.S \end{align*}

Show that: $\cot (\alpha-\beta)=\frac{\cot \alpha \cot \beta+1}{\cot \beta-\cot \alpha}$

Solution.

\begin{align*} L.H.S & = \cot (\alpha-\beta) \\ & = \frac{\cos (\alpha-\beta)}{\sin (\alpha-\beta)} \\ & = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} \\ & = \frac{\frac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta} + \frac{\sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\frac{\sin \alpha \cos \beta}{\sin \alpha \sin \beta} - \frac{\cos \alpha \sin \beta}{\sin \alpha \sin \beta}} \\ & = \frac{\sin \alpha \sin \beta(\cot \alpha \cot \beta + 1)}{ \sin \alpha \sin \beta(\cot \beta- \cot \alpha} )\\ & = \frac{\cot \alpha \cot \beta + 1}{ \cot \beta- \cot \alpha} \\ & = R.H.S \end{align*}

Show that: $\frac{\cos 4 \theta}{\csc \theta}+\frac{\sin 4 \theta}{\sec \theta}=\sin 5 \theta$

Solution.

\begin{align*} L.H.S & = \frac{\cos 4 \theta}{\csc \theta} + \frac{\sin 4 \theta}{\sec \theta} \\ & = \cos 4 \theta \cdot \sin \theta + \sin 4 \theta \cdot \cos \theta \\ & = \sin(4\theta + \theta) \\ & = \sin 5\theta \\ & = R.H.S \end{align*}

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