Question 15 Marks
Find the maximum and minimum values of each of the following trigonometrical expressions:
$\sin\text{x}-\cos\text{x}+1$
AnswerLet $\text{f}(\theta)=\sin\theta-\cos\theta+1$
$\text{f}(\theta)=\sin\theta+(-1)\cos\theta+1$
$(-1)\cos\theta+\sin\theta+1$
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.
View full question & answer→Question 25 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.
View full question & answer→Question 35 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.
View full question & answer→Question 45 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}.$
View full question & answer→Question 55 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).$
AnswerWe 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$]$
$\because\tan(2\alpha+\beta)=\frac{-1}{\sqrt{3}}$
View full question & answer→Question 65 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.
View full question & answer→Question 75 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 $\sin\theta=\frac{\lambda+1}{\lambda-1}\sin\phi$
AnswerLet $\alpha\text{ and }\beta$ 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$\frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\beta}{\cos\beta}}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\frac{\sin\alpha\cos\beta+ \cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}=\frac{\lambda+1}{\lambda-1}$
$\Rightarrow\frac{\sin\theta}{\sin\phi}=\frac{\lambda+1}{\lambda-1}$ $(\theta=\alpha+\beta\text{ and }\phi=\alpha-\beta)$
$\sin\theta=\frac{\lambda+1}{\lambda-1}\sin\phi$
View full question & answer→Question 85 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.
View full question & answer→Question 95 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.
View full question & answer→Question 105 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. $
AnswerWe 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.
View full question & answer→Question 115 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.
View full question & answer→Question 125 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$
AnswerWe 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.
View full question & answer→Question 135 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)$
View full question & answer→Question 145 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$
Answer$\text{L.H.S}=\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$
$=\frac{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{B}\cos\text{C}-\cos\text{B}\sin\text{C}}{\cos\text{B}\cos\text{C}}\\\ \ +\frac{\sin\text{C}\cos\text{A}-\cos\text{C}\sin\text{A}}{\cos\text{C}\cos\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}-\frac{\cos\text{A}\sin\text{A}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{B}\cos\text{C}}{\cos\text{B}\cos\text{C}}\\\ \ -\frac{\cos\text{B}\sin\text{C}}{\cos\text{B}\cos\text{C}}+\frac{\sin\text{C}\cos\text{A}}{\cos\text{C}\cos\text{A}}-\frac{\cos\text{C}\sin\text{A}}{\cos\text{C}\cos\text{A}}$
$$$=\tan\text{A}-\tan\text{B}+\tan\text{B}-\tan\text{C}+\tan\text{C}-\tan\text{A}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.
View full question & answer→Question 155 Marks
Reduce each of the following expressions to the sine and cosin of a single expression:
$24\cos\text{x}+7\sin\text{x}$
AnswerLet $\text{f(x)}=24\cos\text{x}+7\sin\text{x}$
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,
$\text{f(x)}=25\Big(\frac{24}{25}\cos\text{x}+\frac{7}{25}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=25\Big(\cos\alpha\cos\text{x}+\sin\alpha\sin\text{x}\Big),$ $\text{where}\cos\alpha=\frac{24}{25}\text{ and}\sin\alpha=\frac{7}{25}.$
$\Rightarrow\text{f(x)}=25\cos\Big(\alpha-\text{x}\Big), \text{where}\tan\alpha=\frac{24}{7}.$
View full question & answer→Question 165 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.
View full question & answer→Question 175 Marks
If $ \tan\text{A}+\tan\text{B}=\text{a}$ and $\cot\text{A}+\cot\text{B}=\text{b},$ prove that $\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}.$
AnswerWe have,
$\tan\text{A}+\tan\text{B}=\text{a}$ $\text{and}\ \cot\text{A}+\cot\text{B}=\text{b}$
$\cot\text{A}+\cot\text{B}=\text{b}$
$\Rightarrow\frac{1}{\tan\text{A}}+\frac{1}{\tan\text{B}}=\text{b}$ $\Big[\because\cot\theta=\frac{1}{\tan\theta}\Big]$
$\Rightarrow\frac{\tan\text{B}+\tan\text{A}}{\tan\text{A}\tan\text{B}}=\text{b}$ $\big[\because\tan\text{A}+\tan\text{B}=\text{a}\big]$
$\Rightarrow\frac{\text{a}}{\tan\text{A}\tan\text{B}}=\text{b}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\tan\text{A}\tan\text{B}$
$\because\cot\text{(A}+\text{B)}=\frac{1}{\tan\text{(A}+\text{B)}}$
$=\frac{\frac{1}{\tan\text{A}+\tan\text{B}}}{1-\tan\text{A}\tan\text{B}}$
$=\frac{1-\tan\text{A}\tan\text{B}}{\tan\text{A} +\tan\text{B}}$
$=\frac{1-\frac{\text{a}}{\text{b}}}{\text{a}}$
$=\frac{\text{b}-\text{a}}{\text{ab}}$ $\big[\because\tan\text{A}+\tan\text{B}=\frac{\text{a}}{\text{b}}\big]$
$=\frac{\text{b}}{\text{ab}}-\frac{\text{a}}{\text{ab}}$
$=\frac{\text{1}}{\text{a}}-\frac{\text{1}}{\text{b}}$
$\therefore\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}$
Hence proved.
View full question & answer→Question 185 Marks
If $\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$ prove that $1+\cot\alpha\tan\beta$
AnswerWe have,
$\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$
$\Rightarrow-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=-1$
$\Rightarrow-\cos(\alpha+\beta)=1\ ...(1)$
$\therefore\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$=\sqrt{1-1^2}=0$
$\Rightarrow\sin(\alpha+\beta)=0\ ...(2)$
Now,
$1+\cot\alpha\tan\beta=1+\frac{\cos\alpha}{\sin\alpha}\times\frac{\sin\beta}{\cos\beta}$
$=\frac{\sin\alpha\times\cos\beta+\cos\alpha\times\sin\beta}{\sin\alpha\times\cos\beta}$
$=\frac{\sin(\alpha+\beta)}{\sin\alpha\times\cos\beta}=\frac{0}{\sin\alpha\times\cos\beta}$ [Using equation (2)]
$\therefore1+\cot\alpha\tan\beta=0$
Hence proved.
View full question & answer→Question 195 Marks
Reduce each of the following expressions to the sine and cosin of a single expression:
$\cos\text{x}-\sin\text{x}$
AnswerLet $\text{f(x)}=\cos\text{x}-\sin\text{x}$
Dividing and multiplying by $\sqrt{1^2+1^2},$ i.e. by $\sqrt{2},$ we get:
$\text{f(x)}=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\cos\text{x}-\frac{1}{\sqrt{2}}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=\sqrt{2}(\cos45^\circ\cos\text{x}-\sin45^\circ\sin\text{x})$
$\Rightarrow\text{f(x)}=\sqrt{2}\cos\Big(\frac{\pi}{4}+\text{x}\Big)$
Again,
$\text{f(x)}=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\cos\text{x}-\frac{1}{\sqrt{2}}\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=\sqrt{2}\Big(\sin45^\circ\cos\text{x}-\cos45^\circ\sin\text{x}\Big)$
$\Rightarrow\text{f(x)}=\sqrt{2}\sin\Big(\frac\pi4-\text{x}\Big)$
View full question & answer→Question 205 Marks
Show that $\sin100^\circ-\sin10^\circ$ is positive.
AnswerWe have, $\sin100^\circ-\sin10^\circ$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\times100^\circ-\frac{1}{\sqrt{2}}\times\cos100^\circ\Big)$ $\big[$Multiplying and dividing by $\sqrt{1^2+1^2}$ i.e., by $\sqrt{2}\big]$
$=\sqrt{2}(\cos45^\circ\times\sin100^\circ-\sin45^\circ\times\cos100^\circ)$
$=\sqrt{2}(\sin100^\circ\times\cos45^\circ-\cos100^\circ\times\sin45^\circ)$
$=\sqrt{2}(\sin(100^\circ-45^\circ))$
$=\sqrt{2}\sin55^\circ,$ which is positive real number. $[\because\sin\theta$ is positive in first quadrant$]$
View full question & answer→Question 215 Marks
If are two different valus of X lying between 0 and which satisfy the equation $6\cos\text{x}+8\sin\text{x}=9$find the value of $\sin(\alpha+\beta).$
AnswerWe have,
$6\cos\text{x}+8\sin\text{x}=9\ ...(1)$
$\Rightarrow8\sin\text{x}=9-6\cos\text{x}$
$\Rightarrow\big(8\sin\text{x}\big)^2=\big(9-6\cos\text{x}\big)^2$
$\Rightarrow64\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow64\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow64(1-\cos^2\text{x})=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow64-64\cos^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$
$\Rightarrow36\cos^2\text{x}+64\cos^2\text{x}-108\cos\text{x}-108\cos\text{x}$
$\Rightarrow100\cos^2\text{x}-108\cos\text{x}+17=0\ ...(2)$
Since $\alpha,\beta$ are roots of equation ...(i)
$$$\therefore\cos\alpha$ and $\cos\beta$ roots of equation ....(ii)
$\because\cos\alpha+\cos\beta=\frac{17}{100}\ ...(3)$
$\text{Again},6\cos\text{x}+8\sin\text{x}=9$
$\Rightarrow6\cos\text{x}=9-8\sin\text{x}$
$\Rightarrow\big(8\cos\text{x}\big)^2=\big(9-6\sin\text{x}\big)^2$
$\Rightarrow36\cos^2\text{x}=81+64\cos^2\text{x}-144\cos\text{x}$
$\Rightarrow36(1-\sin^2\text{x})=81+64\sin^2\text{x}-144\sin\text{x}$
$\Rightarrow36-36\sin^2\text{x}=81+36\cos^2\text{x}-108\cos\text{x}$ $[\because$ squaring both sides$]$
$\Rightarrow64\sin^2\text{x}+36\sin^2\text{x}-144\sin\text{x}+81-36=0$
$\Rightarrow100\sin^2\text{x}-144\sin\text{x}+45=0\ ...(4)$
$\because\sin\alpha\times\sin\beta=\frac{45}{100}\ ...(5)$
Now, $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$=\frac{17}{100}-\frac{45}{100}$$=-\frac{28}{100}$$=-\frac{7}{25}$ [Using equation (3) and (5)]
Now, $\sin(\alpha+\beta)=\sqrt{1-(\cos\text{x})^2}$
$=\sqrt{1-\Big(-\frac{7}{25}\Big)^2}$$=\sqrt{1-\frac{49}{625}}$ $=\sqrt{\frac{625-49}{625}}$
$=\sqrt{\frac{576}{625}}$$=\frac{24}{25}$ $\therefore\sin(\alpha+\beta)=\frac{24}{25}$
View full question & answer→Question 225 Marks
Reduce each of the following expressions to the sine and cosin of a single expression:
$\sqrt{3}\sin\text{x}-\cos\text{x}$
AnswerLet $\text{f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}$
Dividing and multiplying by $\sqrt{3+1},$ i.e. by 2, We get:
$\text{f(x)}=2\Big(\frac{\sqrt{3}}{2}\sin\text{x}-\frac{1}{2}\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=2\Big(\cos\frac\pi6\sin\text{x}-\sin\frac\pi6\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=2\sin\Big(\text{x}-\frac{\pi}{6}\Big)$
Again,
$\text{f(x)}=2\Big(\frac{\sqrt{3}}{2}\sin\text{x}-\frac{1}{2}\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=2\Big(\sin\frac{\pi}{3}\sin\text{x}-\cos\frac\pi3\cos\text{x}\Big)$
$\Rightarrow\text{f(x)}=-2\cos\Big(\frac\pi3+\text{x}\Big)$
View full question & answer→Question 235 Marks
Prove that $(2\sqrt{3}+3)\sin\text{x}+2\sqrt{3}\cos\text{x}$ lies between $-(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15}).$
AnswerLet $\text{f(x)}=(2\sqrt{3}+3)\sin\text{x}+2\sqrt{3}\cos\text{x}$
We know that,
$-\sqrt{(2\sqrt{3}+3)^2+(2\sqrt{3})^2}\le\text{f(x)}\le\sqrt{(2\sqrt{3}+3)^2+(2\sqrt{3})^2}$
$\Rightarrow-\sqrt{12+9+12\sqrt{3}+12}\le\text{f(x)}\le\sqrt{12+9+12\sqrt{3}+12}$
$\Rightarrow-\sqrt{33+12\sqrt{3}}\le\text{f(x)}\le\sqrt{33+12\sqrt{3}}$
Disclaimer: Instead of $-(2\sqrt{3}+\sqrt{15})$ and $(2\sqrt{3}+\sqrt{15}),$ it should be $-\sqrt{33+12\sqrt{3}}$ and $\sqrt{33+12\sqrt{3}}.$
View full question & answer→Question 245 Marks
If X lies in the first quadrant and $\cos\text{x}=\frac{8}{17},$ then prove that
$\cos\Big(\frac{\pi}{6}+\text{x}\Big) +\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{2\pi}{3}-\text{x}\Big)=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\frac{23}{17}$
AnswerWe have,
$\cos\text{x}=\frac{8}{17}$
$\therefore\sin\text{x}=\sqrt{1-\cos^2\text{x}}=\sqrt{1-\frac{64}{289}}$
$=\sqrt{\frac{225}{289}}$
$=\frac{15}{17}$
Now, $\cos\Big(\frac{\pi}{6}+\text{x}\Big) +\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{2\pi}{3}-\text{x}\Big)$
$=\Big[\cos\frac{\pi}{6}\cos\text{x}-\sin\frac{\pi}{6}\sin\text{x}\Big]+\Big[\cos\frac{\pi}{4}\cos\text{x}+\sin\frac{\pi}{4}\sin\text{x}\Big]\\\ \ +\Big[\cos\frac{\pi}{4}\cos\text{x}+\sin\frac{\pi}{4}\sin\text{x}\Big]$
$$$=\Big[\cos\frac{\pi}{6}+\cos\frac{\pi}{4}+\cos\frac{2\pi}{3}\Big]\cos\text{x}\\\ \ +\sin\text{x}\Big[-\sin\frac{\pi}{6}+\sin\frac{\pi}{4}+\sin\frac{2\pi}{3}\Big]$
$$$=\Big[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}+\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)\Big]\times\frac{8}{17}\\\ +\frac{15}{17}\times\Big[\frac{1}{2}+\frac{1}{\sqrt{2}}+\sin\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)\Big]$
$$$=\Big[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\sin\frac{\pi}{6}\Big]\times\frac{8}{17}+\frac{15}{17}\times\Big[-\frac{1}{2}+\frac{1}{\sqrt{2}} +\cos\frac{\pi}{6}\Big]$
$=\Big[\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\Big]\times\frac{8}{17}+\frac{15}{17}\times\Big[-\frac{1}{2}+\frac{1}{\sqrt{2}} +\frac{\sqrt{3}}{2}\Big]$ $\big[\because\cos\text{A}$ is negative in second quadrant$\big]$
$=\Big[\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big]\times\frac{8}{17}\times\frac{15}{17}\times\Big[\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big]$
$=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\Big(\frac{8}{17}+\frac{15}{17}\Big)$
$=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\times\frac{23}{17}$
$\therefore\cos\Big(\frac{\pi}{6}+\text{x}\Big) +\cos\Big(\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{2\pi}{3}-\text{x}\Big)=\Big(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\Big)\frac{23}{17}$
Hence proved.
View full question & answer→Question 255 Marks
If than $\alpha=\text{x}+1,\tan\beta=\text{x}-1,$ prove that $2\cot(\alpha-\beta)=\text{x}^2$
AnswerWe have,
$\tan\alpha=\text{x}+1\text{ and }\tan\beta=\text{x}-1$
Now, $2\cot(\alpha-\beta)$
$=\frac{2}{\tan(\alpha-\beta)}$
$=\frac{\frac{2}{\tan\alpha-\tan\beta}}{1+\tan\alpha\tan\beta}$
$=\frac{2(1+\tan\alpha\tan\beta)}{\tan\alpha-\tan\beta}$
$=\frac{2[1+(\text{x}+1)(\text{x}-1)]}{\text{x}+1-(\text{x}-1)}$
$=\frac{2[1+\text{x}^2-1]}{\text{x}+1-\text{x}+1}$
$=\frac{2\text{x}\text{x}^2}{2}=\text{x}^2$
$\therefore2+\cot(\alpha+\beta )=\text{x}^2$
Hence proved.
View full question & answer→Question 265 Marks
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ show that
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$
Answer$\text{a}^2+\text{b}^2=(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2$
$\Rightarrow\text{a}^2+\text{b}^2=\sin^2\alpha+\sin^2\beta+2\sin^2\alpha\sin^2\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta$
$\Rightarrow\text{a}^2+\text{b}^2=\sin^2\alpha+\cos^2\alpha+\sin^2\beta+\cos^2\beta+2(\sin\alpha+\sin\beta+\cos\alpha\cos\beta)$
$\Rightarrow\text{a}^2+\text{b}^2=2+2\cos(\alpha-\beta)\cdots(1)$
Now,
$\text{b}^2-\text{a}^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha-\sin\beta)^2$
$\Rightarrow\text{b}^2-\text{a}^2=\cos^2+\cos^2\beta+\sin^2\alpha-\sin^2\beta+2\cos\alpha+\cos\beta-2\sin\alpha\sin\beta$
$\Rightarrow\text{b}^2-\text{a}^2=(\cos^2\alpha-\sin^2\beta)+(\cos^2\beta-\sin^2\alpha)-2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2-\text{a}^2=2\cos(\alpha+\beta)\cos(\alpha-\beta)+2\cos(\alpha-\beta)$
$\Rightarrow\text{b}^2-\text{a}^2=\cos(\alpha+\beta)\Big(2+2\cos(\alpha-\beta)\Big)\cdots(2)$
From (1) and (2), we have
$\text{b}^2-\text{a}^2=\cos(\alpha+\beta)(\text{a}^2+\text{b}^2)$
$\Rightarrow\frac{\text{b}^2-\text{a}^2}{\text{a}^2-\text{b}^2}=\cos(\alpha+\beta)$
$\Rightarrow\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$\Rightarrow\sin(\alpha+\beta)=\sqrt{1-\Big(\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}\Big)^2}=\sqrt{\frac{\text{b}^4+\text{a}^4-\text{b}^4-\text{a}^4+4\text{a}^2\text{b}^2}{(\text{b}^2+\text{a}^2)}}$
$\Rightarrow\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$
View full question & answer→Question 275 Marks
If then $\text{x}+\tan\Big(\frac{\pi}{3}\Big)+\tan\Big(\text{x}=\frac{2\pi}{3}\Big)=3,$prove that $\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}=1$
AnswerWe have,
$\tan\text{x}+\tan\Big(\text{x}+\frac{\pi}{3}\Big)+\tan\Big(\text{x}+\frac{2\pi}{3}\Big)=3$
$\Rightarrow \tan\text{x}+\Bigg[\frac{\tan\text{x}+\tan\frac{\pi}{3}}{1-\tan\text{x}\tan\frac{\pi}{3}}\Bigg]+\Bigg[\frac{\tan\text{x}+\tan\Big(\frac{2\pi}{3}\Big)}{1-\tan\text{x}\tan\frac{2\pi}{3}}\Bigg]=3$
$\Rightarrow \tan\text{x}+\Bigg[\frac{\tan\text{x}+\tan\frac{\pi}{3}}{1-\sqrt{3}\tan\text{x}}\Bigg]+\Bigg[\frac{\tan\text{x}+\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)}{1-\tan\text{x}\tan\Big(\frac{\pi}{2}+\frac{\pi}{3}\Big)}\Bigg]=3$
$\Rightarrow \tan\text{x}+\frac{\tan\text{x}+ \sqrt{3}}{1-\sqrt{3}\tan\text{x}}+\frac{\tan\text{x}-\cot\frac{\pi}{3}}{1+\tan\text{x}\cot\frac{\pi}{3}}=3$
$\Rightarrow \tan\text{x}+\frac{\tan\text{x}+ \sqrt{3}}{1-\sqrt{3}\tan\text{x}}+\frac{\tan\text{x}-\sqrt{3}}{1+\sqrt{3}\tan\text{x}}=3$
$\Rightarrow\tan\text{x}+\frac{\Big(\tan\text{x}+\sqrt{3}\Big)\Big(1+\sqrt{3}\tan\text{x}\Big ) +\Big(\tan\text{x}-\sqrt{3}\Big)\Big(1-\sqrt{3}\tan\text{x}\Big)}{\Big(1-\sqrt{3}\tan\text{x}\Big)\Big(1+ \sqrt{3}\tan\text{x}\Big)}=3$
$\Rightarrow\tan\text{x}+\frac{\tan\text{x}+\sqrt{3}\tan^2\text{x}+\sqrt{3}\tan\text{x} +\tan\text{x}-\sqrt{3}\tan^2\text{x}-\sqrt{3}+3\tan\text{x}}{1-\Big(\sqrt{3}\tan\text{x}\Big)^2}$
$\Rightarrow\tan\text{x}+\frac{8\tan\text{x}}{1-3\tan^2\text{x}}=3$
$\Rightarrow\tan\text{x}+\frac{\Big(1-3\tan^2\text{x}\Big)+8\tan\text{x}}{1-3\tan^2\text{x}}=3$
$\Rightarrow\tan\text{x}-\frac{3\tan^3\text{x}+8\tan\text{x}}{1-3\tan^2\text{x}}=3$
$\Rightarrow\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}=1$
Hence proved.
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