Solve the Following Question.(4 Marks) - Maths STD 11 Science Questions - Vidyadip

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Solve the Following Question.(4 Marks)

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16 questions · timed · auto-graded

Question 14 Marks

Find the maximum and minimum values of each of the following trigonometrical expressions:
$\sin\text{x}-\cos\text{x}+1$

Answer

Let $\text{f}(\theta)=\sin\theta-\cos\theta+1$
$\text{f}(\theta)=\sin\theta+(-1)\cos\theta+1$
$(-1)\cos\theta+\sin\theta+1$
We know that,
$-\sqrt{(-1)^2+(1)^2}\leq\cos\theta+\sin\theta\leq\sqrt{(-1)^2+(1)^2}$
$\Rightarrow-\sqrt{1+1}\leq-\cos\theta+\sin\theta\leq\sqrt{1+1}$
$\Rightarrow-\sqrt{2}-\cos\theta+\sin\theta\leq\sqrt{2}$
$\Rightarrow-\sqrt{2}+1\leq-\cos\theta+\sin\theta+1\leq\sqrt{2}+1$
$\Rightarrow1-\sqrt{2}\leq\text{f}(\theta)\leq1+\sqrt{2}$
Hence, minimum and maximum values of $\sin\theta-\cos\theta+1-\sqrt{2}$ are $1 +\sqrt{2}$ respectively.

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Question 24 Marks

Prove that:$\sin^2\text{B}=\sin^2\text{A}+\sin^2\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\sin\text{(A}-\text{B)} $

Answer

$\text{R.H.S}=\sin^2\text{A}+\sin^2\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\sin\text{(A}-\text{B)} $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[\sin\text{(A}-\text{B)}-2\sin\text{A}\cos\text{B}\Big] $
$=\sin^2\text{A}+\sin\text{(A}-\text{B)}\Big[-\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\Big[\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\ \Big]$
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}(\sin(\text{A}+\text{B))} $
$=\sin^2\text{A}-\sin\text{(A}-\text{B)}\sin(\text{A}+\text{B)} $
$=\sin^2\text{A}-\sin(\sin^2\text{A}-\sin^2\text{B})$
$=\sin^2\text{A}-\sin\sin^2\text{A}+\sin^2\text{B}$$\Big[\because\sin(\text{A}-\text{B)}\sin\text{(A+}\text{B)}=\sin^2\text{A}-\sin^2\text{B}\Big]$
$=\sin^2\text{B}$
$=\text{L.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 34 Marks

Prove that:
$\frac{1}{\cos\text{(x}-\text{b})\cos\text{(x}-\text{b)}}=\frac{\tan\text{(x}-\text{a)}-\tan\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$

Answer

$\text{L.H.S}=\frac{1}{\cos\text{(x}-\text{a)}\cos\text{(x}-\text{b)}}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(a}-\text{b)}}{\cos\text{(x}-\text{b})\cos\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)-}(\text{x}-\text{b)}}{\cos\text{(x}-\text{a})\cos\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)cos}(\text{x}-\text{a)}-\cos\text{(x}-\text{b})\sin\text{(x}-\text{a)}}{\cos\text{(x}-\text{a})\cos\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b})}\Big[\frac{\sin\text{(x}-\text{b)}\cos\text{(x}-\text{a)}}{\cos\text{(x}-\text{b)}\cos\text{(x}-\text{b})}-\frac{\cos\text{(x}-\text{b)}\sin\text{(x}-\text{a)}}{\cos\text{(x}-\text{b)}\cos\text{(x}-\text{b})}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b})}\Big[\frac{\sin\text{(x}-\text{b)}}{\cos\text{(x}-\text{b)}}-\frac{\sin\text{(x}-\text{a)}}{\cos\text{(x}-\text{b})}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b})}\Big[\tan\text{(x}-\text{b)}-\tan\text{(x}-\text{b)}\Big]$
$=\frac{\tan\text{(x}-\text{b)-}\tan\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$
$=\text{R.H.S}$
$\therefore\text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 44 Marks

If $\alpha\text{ and }\beta$ are two solutions of the equation $\text{a}\tan\text{x+b}\sec\text{x}=\text{c},$ the find the value of $\sin(\alpha+\beta)\text{ and }\cos(\alpha+\beta).$

Answer

$\text{a}\tan\text{x}+\text{b}\sec\text{x}=\text{c}$
$\Rightarrow(\text{c}-\text{a}\tan\text{x})=\text{b}\sec\text{x}$
$\Rightarrow(\text{c}-\text{a}\tan\text{x})^2=(\text{b}\sec\text{x})^2$
$\Rightarrow\text{c}^2+\text{a}^2\tan^2\text{x}-2\text{a}\text{c}\tan\text{x}=\text{b}^2\sec^2\text{x}$
$\Rightarrow\text{c}^2+\text{a}^2\tan^2\text{x}-2\text{a}\text{c}\tan\text{x}=\text{b}^2(1+\tan^2\text{x)}$
$\Rightarrow(\text{a}^2-\text{b}^2)\tan^2\text{x}-2\text{a}\text{c}\tan\text{x}+\text(\text{c}^2-\text{b}^2)=0$
This is a quadratic in $\tan\text{x}.$
It has two solutions $\tan\alpha\text{ and }\tan\beta.$
$\tan\alpha\text{ and }\tan\beta=\frac{2\text{ac}}{\text{a}^2-\text{b}^2}$
$\tan\alpha\times\tan\beta=\frac{\text{c}^2-\text{b}^2}{\text{a}^2-\text{b}^2}$
$\therefore\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$=\frac{\frac{2\text{ac}}{\text{a}^2-\text{b}^2}}{1-\frac{\text{c}^2-\text{b}^2}{\text{a}^2-\text{b}^2}}=\frac{2\text{ac}}{\text{a}^2-\text{c}^2}$
Hence, $\sin(\alpha+\beta)=\frac{ 2\text{ac}}{\text{a}^2-\text{b}^2}\text{ and }\cos(\alpha+\beta)=\frac{\text{a}^2-\text{c}^2}{\text{a}^2+\text{b}^2}.$

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Question 54 Marks

If $\sin(\alpha+\beta)=1$ and $\sin(\alpha-\beta)=\frac{1}{2}$ where $0\leq\alpha,\beta\leq\frac{\pi}{2}$ than find the values of $\tan(\alpha+2\beta)$and $\tan(2\alpha+\beta).$

Answer

We have,
$\sin(\alpha+\beta)=1$
$\Rightarrow\sin(\alpha-\beta)=\frac{\pi}{2}$
$\Rightarrow\alpha+\beta=\frac{\pi}{2}\ ...(1)$
and, $\sin(\alpha-\beta)=\frac{1}{2}$
$\Rightarrow\sin(\alpha-\beta)=\sin\frac{\pi}{6}$
$\Rightarrow\alpha-\beta=\frac{\pi}{6}\ ...(2)$
Adding equations (1) and (2), we get
$2\alpha=\frac{\pi}{2}+\frac{\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
Putting in equation (1), we get
$\frac{\pi}{3}+\beta =\frac{\pi}{2}$
$\Rightarrow\beta=\frac{\pi}{2}-\frac{\pi}{2}$$\Rightarrow \beta=\frac{3\pi-2\pi}{6}$$=\frac{\pi}{6}$
$\Rightarrow\beta=\frac{\pi}{6}$
Now, $\tan(\alpha+2\beta)=\tan\Big(\frac{\pi}{3}+2\times\frac{\pi}{6}\Big)$
$=\tan\Big(\frac{\pi}{3}+\frac{\pi}{3}\Big)$$=-\cot\frac{\pi}{6}$$=-\sqrt{3}$$[\because\tan\theta $ negative in second quadrant and, $\tan(2\alpha+\beta)=\tan\Big(2\times\frac{\pi}{3}+\frac{\pi}{3}\Big)$
$=\tan\Big(\frac{2\pi}{3}+\frac{\pi}{6}\Big)$$=\tan\Big(\frac{4\pi+\pi}{6}\Big)$$=\tan\Big(\frac{5\pi}{6}\Big)$
$=\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)$$=-\cot\frac{\pi}{3}$$[\because\tan\theta$ is negative in second quadrant$]$
$\because\tan(2\alpha+\beta)=\frac{-1}{\sqrt{3}}$

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Question 64 Marks

Prove that: $\frac{\tan\text{(A}+\text{B)}}{\cot\text{(A}-\text{B)}}=\frac{\tan^2\text{A}-\tan^2\text{B}}{1-\tan^2\text{A}\tan^2\text{B}}$

Answer

$\text{L.H.S}=\frac{\tan\text{(A}+\text{B)}}{\cot\text{(A}-\text{B)}}$
$=\frac{\frac{\tan\text{(A}+\text{B)}}{1}}{\tan\text{(A}-\text{B)}}$
$=\tan\text{(A}+\text{B)}\tan\text{(A}-\text{B)}$
$=\Big[\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]\Big[\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}\Big]$
$=\frac{(\tan\text{A}+\tan\text{B)}(\tan\text{A}-\tan\text{B)}}{(1-\tan\text{A}\tan\text{B)}(1+\tan\text{A}\tan\text{B)}}$
$=\frac{\tan^2\text{A}-\tan^2\text{B}}{1-(\tan\text{A}\tan\text{B)}^2}$
$=\frac{\tan^2\text{A}-\tan^2\text{B}}{1-\tan^2\text{A}\tan^2\text{B}}$ $\big[\because\text{(a}-\text{b)}\text{(a}+\text{b)}=\text{a}^2-\text{b}^2\big]$
$=\text{R.H.S}$
$\therefore\text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 74 Marks

If angle $\theta$ is divided into two parts such that the tangents of one part is $\lambda$ times the tangent of other,and $\phi$ is their difference,the show that $\sin\theta=\frac{\lambda+1}{\lambda-1}\sin\phi$

Answer

Let $\alpha\text{ and }\beta$ be the two parts of angle $\theta$ then,$\theta=\alpha+\beta\text{ and }\phi=\alpha-\beta$
Now,$\tan\alpha=\lambda\tan\beta$
$\Rightarrow\frac{\tan\alpha}{\tan\beta}=\frac{\lambda}{1}$
Applying componendo and dividendo, we get$\frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\beta}{\cos\beta}}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\frac{\sin\alpha\cos\beta+ \cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\sin\theta}{\sin\phi}=\frac{\lambda+1}{\lambda-1}$ $(\theta=\alpha+\beta\text{ and }\phi=\alpha-\beta)$
$\sin\theta=\frac{\lambda+1}{\lambda-1}\sin\phi$

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Question 84 Marks

If $\tan\text{x}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha},$ then show that $\sin\alpha+\cos\alpha=\sqrt{2}\cos\text{x}.$

Answer

$\tan\text{x}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$
$\Rightarrow\tan\text{x}=\frac{\tan\alpha-1}{\tan\alpha+1}$ [Dividing both numerator and denominator by $\cos\alpha$]
$\Rightarrow\tan\theta=\frac{\tan\alpha-\tan\frac{\pi}{4}}{1+ \tan\frac{\pi}{4}.\tan\alpha}$
$\Rightarrow\tan\theta=\tan\Big(\alpha-\frac{\pi}{4}\Big)$
$\Rightarrow\theta=\alpha-\frac{\pi}{4}$ [Removing tan form both sides]
$\Rightarrow\cos\theta=\cos\Big(\alpha-\frac{\pi}{4}\Big)$ [Taking cos on both sides]
$\Rightarrow\cos\theta=\cos\alpha.\cos\frac{\pi}{4}+\sin\alpha.\sin\frac{\pi}{4}$
$\Rightarrow\cos\theta=\cos\alpha.\frac{1}{\sqrt{2}}+\sin\alpha.\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\theta=\frac{\cos\alpha+\sin\alpha}{\sqrt{2}}$
$\Rightarrow\sqrt{2}\cos\theta=\sin\alpha+\cos\alpha$
Hence proved.

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Question 94 Marks

Prove that:
$\frac{1}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}=\frac{\cot\text{(x}-\text{b)}-\cot\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$

Answer

$\text{L.H.S}\ \frac{1}{\sin\text{(x}-\text{a)}\sin\text{(x}-\text{b)}}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(a}-\text{b)}}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)-}(\text{x}-\text{b)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin\text{(x}-\text{b)}\cos(\text{x}-\text{a})-\cos(\text{x}-\text{b})\sin(\text{x}-\text{a)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\frac{\sin(\text{x}-\text{b)}\cos(\text{x}-\text{a})}{\sin\text{(x}-\text{a)}\sin(\text{x}-\text{b})}-\frac{\sin(\text{x}-\text{b)}\cos(\text{x}-\text{a})}{\sin\text{(x}-\text{a)}\sin(\text{x}-\text{b})}\Big]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\Big[\cot(\text {x}-\text{a}-\cot(\text{x}-\text{b})\Big]$
$=\frac{\cot(\text{x}-\text{a})-\cot(\text{x}-\text{b})}{\sin(\text{a}-\text{b})}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 104 Marks

$\text{If} \ \cos\text{A}+\sin\text{B}=\text{m}$ and $\sin\text{A}+\cos\text{B}=\text{n},$ prove that $2\sin\text{(A}+\text{B)}=\text{m}^2+\text{n}^2-2. $

Answer

We have,
$\cos\text{A}+\sin\text{B}=\text{m}\ \text{and}\ \sin\text{A}+\cos\text{B}=\text{n}$
$=\text{m}^2+\text{n}^2-2$
$=(\cos\text{A}+\sin\text{B})^2+(\sin\text{A}+\cos\text{B)}^2-2$
$ =\cos^2\text{A}+\sin^2\text{B}+2\cos\text{A}\sin\text{B}+\sin^2\text{A}\cos^2\text{B}+2\sin\text{A}\cos\text{B}-2$
$ =(\sin^2\text{A}+\cos^2\text{A})+(\sin^2\text{B}+\cos^2\text{B})\\\ +2\cos\text{A}\sin\text{B}+\sin^2\text{A}\cos \text{B}+2\sin\text{A}\cos\text{B}-2 $
$=1+1+2\cos\text{A}\sin\text{B}+2\sin\text{A}\cos\text{B}-2$
$=2+2(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B)}-2$
$=2(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B)}$
$ =2\sin\text{(A}+\text{B)}$$\big[\because\sin\text{(A}+\text{B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\big]$
$\therefore2\sin\text{(A}+\text{B)}=\text{m}^2+\text{n}^2-2 $
Hence proved.

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Question 114 Marks

Prove that:
$\frac{1}{\sin\text{(x}-\text{b})\sin\text{(x}-\text{b)}}=\frac{\cot\text{(x}-\text{a)}+\tan\text{(x}-\text{b)}}{\cos\text{(a}-\text{b)}}$

Answer

$\text{L.H.S}=\frac{1}{\sin\text{(x}-\text{a)}\cos\text{(x}-\text{b)}}$
$=\frac{1}{\cos\text{(a}-\text{b)}}\Big[\frac{\sin\text{(a}-\text{b)}}{\sin\text{(x}-\text{b})\cos\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\cos\text{(a}-\text{b)}}\Big[\frac{\cos\text{(x}-\text{b)-}(\text{x}-\text{b)}}{\sin\text{(x}-\text{a})\cos\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\cos\text{(a}-\text{b)}}\Big[\frac{\cos\text{(x}-\text{b)cos}(\text{x}-\text{a)}+\sin\text{(x}-\text{b})\sin\text{(x}-\text{a)}}{\sin\text{(x}-\text{a})\sin\text{(x}-\text{b)}}\Big]$
$=\frac{1}{\cos\text{(a}-\text{b})}\Big[\frac{\cos\text{(x}-\text{b)}\cos\text{(x}-\text{a)}}{\sin\text{(x}-\text{b)}\cos\text{(x}-\text{b})}+\frac{\sin\text{(x}-\text{b)}\sin\text{(x}-\text{a)}}{\sin\text{(x}-\text{b)}\cos\text{(x}-\text{b})}\Big]$
$=\frac{\cot\text{(a}-\text{b)}+\tan\text{(x}-\text{b)}}{\cos\text{(a}-\text{b)}}$
$=\frac{1}{\sin\text{(a}-\text{b})}\Big[\cot\text{(x}-\text{b)}-\cot\text{(x}-\text{b)}\Big]$
$=\frac{\cot\text{(x}-\text{a)-}\cot\text{(x}-\text{b)}}{\sin\text{(a}-\text{b)}}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 124 Marks

Prove that: $\frac{\sin\text{(A-B)}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{(B-C)}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{(C-A)}}{\sin\text{C}\sin\text{A}}=0$

Answer

We have,
$\text{L.H.S }\frac{\sin\text{(A-B)}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{(B-C)}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{(C-A)}}{\sin\text{C}\sin\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{B}\cos\text{C}-\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}\\\ \ +\frac{\sin\text{C}\cos\text{A}-\cos\text{C}\sin\text{A}}{\sin\text{C}\sin\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}}{\sin\text{A}\sin\text{B}}-\frac{\cos\text{A}\sin\text{A}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{B}\cos\text{C}}{\sin\text{B}\sin\text{C}}\\\ \ -\frac{\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{C}\cos\text{A}}{\sin\text{C}\sin\text{A}}-\frac{\cos\text{C}\sin\text{A}}{\sin\text{C}\sin\text{A}}$
$=\frac{\cos\text{B}}{\sin\text{B}}-\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{C}}{\sin\text{C}}-\frac{\cos\text{B}}{\sin\text{B}}+\frac{\cos\text{A}}{\sin\text{A}}-\frac{\cos\text{C}}{\sin\text{C}}$ $\Big[\because\cot\theta=\frac{\cos\theta}{\sin\theta}\Big]$
$=\cot\text{B}-\cot\text{A}+\cot\text{C}-\cot\text{B}+\cot\text{A}-\cot\text{C}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 134 Marks

If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ show that
$\cos(\alpha+\beta)=\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}$

Answer

$\text{a}^2+\text{b}^2=(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2$
$=\sin^2\alpha+\sin^2\beta+\cos^2\alpha+\cos^2\beta+2\sin\alpha\sin\beta+2\cos\alpha\cos\beta$
$=2+2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2+\text{a}^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha+\sin\beta)^2$
$\Rightarrow\text{b}^2+\text{a}^2=\cos^2\alpha+\cos^2\beta-\sin^2\alpha+\sin^2\beta+2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$
$\Rightarrow\text{b}^2+\text{a}^2=\Big(\cos^2\alpha+\sin^2\beta\Big)+\Big(\cos^2\beta-\sin^2\alpha\Big)-2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2+\text{a}^2=2\cos(\alpha+\beta)+\cos(\alpha-\beta)+2\cos(\alpha-\beta)$
$\Rightarrow\text{b}^2+\text{a}^2=\cos(\alpha+\beta)(2+2\cos(\alpha-\beta) )$
$\Rightarrow\text{b}^2+\text{a}^2=\cos(\alpha+\beta)(\text{a}^2+\text{b}^2 )$
$\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}=\cos(\alpha+\beta)$

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Question 144 Marks

Prove that: $\frac{\sin\text{(A-B)}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{(B-C)}}{\cos\text{B}\cos\text{C}}+\frac{\sin\text{(C-A)}}{\cos\text{C}\cos\text{A}}=0$

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Question 154 Marks

Reduce each of the following expressions to the sine and cosin of a single expression:
$24\cos\text{x}+7\sin\text{x}$

Answer

Let $\text{f(x)}=24\cos\text{x}+7\sin\text{x}$
Dividing and multiplying by $\sqrt{24^2+7^2},$ i.e. by 25, We get:
$\text{f(x)}=25\Big(\frac{24}{25}\cos\text{x}+\frac{7}{25}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=25\Big(\sin\alpha\cos\text{x}+\cos\alpha\sin\text{x}\Big),$ $\text{where}\sin\alpha=\frac{24}{25}\text{ and}\cos\alpha=\frac{7}{25}$
$\Rightarrow\text{f(x)}=25\sin\Big(\alpha+\text{x}\Big), \text{where}\tan\alpha=\frac{24}{7}.$
Again,
$\text{f(x)}=25\Big(\frac{24}{25}\cos\text{x}+\frac{7}{25}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=25\Big(\cos\alpha\cos\text{x}+\sin\alpha\sin\text{x}\Big),$ $\text{where}\cos\alpha=\frac{24}{25}\text{ and}\sin\alpha=\frac{7}{25}.$
$\Rightarrow\text{f(x)}=25\cos\Big(\alpha-\text{x}\Big), \text{where}\tan\alpha=\frac{24}{7}.$

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Question 164 Marks

Prove that:$\cos^2\text{A}+\cos^2\text{B}-2\cos\text{A}\cos\text{B}\cos\text{(A}-\text{B)}=\sin^2\text{(A}+\text{B)} $

Answer

$\text{R.H.S}=\cos^2\text{A}+\cos^2\text{B}-2\cos\text{A}\cos\text{B}\cos\text{(A}+\text{B)}$
$=\cos^2\text{A}+(\text{1}-\sin^2\text{B})-2\cos\text{A}\cos\text{B}\cos(\text{A}+\text{B})$
$=\Big[\cos^2\text{A}-\sin^2\text{B}\Big]-2\cos\text{A}\cos\text{B}\cos\text{(A}+\text{B)}+1$
$=\Big[\cos\text{(A}+\text{B)}\cos\text{(A}-\text{B)}\Big]-2\cos\text{A}\cos\text{B}+\cos\text{(A}+\text{B}) +1$
$=\cos\text{(A}+\text{B)}\Big[\cos(\text{A}-\text{B)}-2\cos\text{A}\cos\text{B}\Big] +1$
$=\cos\text{(A}+\text{B)}\Big[\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}-2\cos\text{A}\cos\text{B}\Big]+1$
$=\cos\text{(A}+\text{B)}\Big[-\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}\Big]+1$
$=\cos\text{(A}+\text{B)}\Big[\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\Big]+1$
$ =\cos^2\text{(A}+\text{B)}+1$
$=1-\cos^2\text{(A}+\text{B)}$ $\Big[\sin^2\theta=1-\cos^2\theta\Big]$
$=\sin^2\text{(A}+\text{B)}$
$=\text{R.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.

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