2 Marks Questions

Question 12 Marks

Prove that:$\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}=-1$

Answer

$\text{L.H.S}=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$
$=\tan(69^\circ+66^\circ)$ $\Big[\because\tan\text{(A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$=\tan(135^\circ)$
$=\tan(90^\circ+45^\circ)$$[\because\tan\theta$ is negetive in second quadrant$]$
$=-\cot45^\circ$
$=-1$
$=\text{R.H.S}$
$\therefore\text{L.H.S}=\text{R.H.S}$
Hence proved.

View full question & answer→

Question 22 Marks

If $\frac{\sin(\text{x}+\text{y})}{\sin(\text{x}-\text{y)}}=\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$ show that $\frac{\tan\text{x}}{\tan\text{y}}=\frac{\text{a}}{\text{b}}.$

Answer

$\frac{\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x}}{\sin\text{x}.\cos\text{y}-\sin\text{y}.\cos\text{x}}=\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$
$\Rightarrow\frac{\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x}+\sin\text{x}.\cos\text{y}-\sin\text{y}.\cos\text{x}}{\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x} -\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x}}$
$=\frac{\text{a}+\text{b}+\text{a}-\text{b}}{\text{a}+\text{b}-\text{a}+\text{b}}$
$\Rightarrow\frac{2\sin\text{x}.\cos\text{y}}{2\sin\text{y}.\cos\text{x}}=\frac{\text{2a}}{\text{2b}}$
$\Rightarrow\frac{\tan\text{x}}{\tan\text{y}}=\frac{\text{a}}{\text{b}}$
Hence proved.

View full question & answer→

Question 32 Marks

Prove that: $\tan\frac{\pi}{12}+\tan\frac{\pi}{6}+\tan\frac{\pi}{12}\tan\frac{\pi}{6}=1$

Answer

We have,
$45^\circ=30^\circ+15^\circ$
$\Rightarrow\tan45^\circ=\tan(30^\circ+15^\circ)$
$\Rightarrow1=\frac{\tan30^\circ+\tan15^\circ}{1-\tan30^\circ\tan15^\circ}$ $\Big[\because\tan\text{(A}+\text{B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$\Rightarrow 1-\tan30^\circ\tan30^\circ=\tan15^\circ+\tan30^\circ$
$\Rightarrow 1=\tan15^\circ+\tan30^\circ+\tan30^\circ\tan15^\circ$
$\Rightarrow \tan15^\circ+\tan30^\circ+\tan15^\circ\tan30^\circ=1$
Hence proved.

View full question & answer→

Question 42 Marks

Prove that: $\cos\frac{7\pi}{12}+\cos\frac{\pi}{12}=\sin\frac{5\pi}{12}-\sin\frac{\pi}{12}$

Answer

$\text{L.H.S}=$ $\cos105^\circ+\cos15^\circ$
$=\cos(90^\circ+15^\circ)+\cos(90^\circ-75^\circ)$
$=-\sin15^\circ+\sin75^\circ$$\Big[\because\cos\big(90+\theta\big)=-\sin\theta\Big]$
$=-\sin75^\circ-\sin15^\circ$$\Big[\cos\big(90-\theta\big)=\sin\theta\Big]$
$=\cos105^\circ+15^\circ=\sin75^\circ-\sin15^\circ$
$=\text{R.H.S}$
$\text{L.H.S}=\text{R.H.S}$
Hence proved.

View full question & answer→

Question 52 Marks

Prove that:
$\sin\Big(\frac{3\pi}{8}-\text{5}\Big)\cos\Big(\frac{\pi}{8}+\text{5}\Big)+\cos\Big(\frac{3\pi}{8}-\text{5}\Big)\sin\Big(\frac{\pi}{8}+\text{5}\Big)=1$

Answer

$\text{L.H.S}$
$\sin\Big(\frac{3\pi}{8}-\text{5}\Big)\cos\Big(\frac{\pi}{8}+\text{5}\Big)+\cos\Big(\frac{3\pi}{8}-\text{5}\Big)\sin\Big(\frac{\pi}{8}+\text{5}\Big)$
$=\sin\Big[\Big(\frac{3\pi}{8}-5\text{}\Big)-\Big(\frac{\pi}{8}+5\text{}\Big)\Big]$ $\big[\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}=\sin\text{(A+B)}\big]$
$=\sin\frac{4\pi}{8} $
$=\sin\frac{\pi}{2}$
$=1$
$\text{R.H.S}$

View full question & answer→

Question 62 Marks

Prove that:
$\sin\Big(\frac{4\pi}{9}+\text{7}\Big)\cos\Big(\frac{\pi}{9}+\text{7}\Big)-\cos\Big(\frac{4\pi}{9}+\text{7}\Big)\sin\Big(\frac{\pi}{9}+\text{7}\Big)=\frac{\sqrt{3}}{2}$$ $

Answer

$\text{L.H.S}$$\sin\Big(\frac{4\pi}{9}+\text{7}\Big)\cos\Big(\frac{\pi}{9}+\text{7}\Big)-\cos\Big(\frac{4\pi}{9}+\text{7}\Big)\sin\Big(\frac{\pi}{9}+\text{7}\Big)$
$=\sin\Big[\Big(\frac{4\pi}{9}+7\text{}\Big)-\Big(\frac{\pi}{9}+7\text{}\Big)\Big]$
$=\sin\frac{3\pi}{9}$ $[\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}=\sin\text{(A - B)}]$
$=\sin\frac{\pi}{3}$
$=\frac{\sqrt{3}}{2}$
$\text{R.H.S}$

View full question & answer→

Question 72 Marks

Prove that:
$\sin\Big(\frac{\pi}{3}-\text{x}\Big)\cos\Big(\frac{\pi}{6=1}+\text{x}\Big)+\cos\Big(\frac{\pi}{3}\text{x}\Big)\sin\Big(\frac{\pi}{6}+\text{x}\Big)=1$

Answer

$\frac{\pi}{3}=60^\circ,\frac{\pi}{6}=30^\circ$ $$$\text{L.H.S}$
$\sin(60^\circ-\text{x})\cos(30^\circ+\text{x)}+\cos(60^\circ-\text{x})\sin(30^\circ+\text{x})$ $=[\sin(60^\circ-\text{x})+(30^\circ+\text{x})]$ $($Using the formula $\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}=\sin\text{(A+B)}$ and taking $\text{A}=60^\circ-\text{x} \text{ and }\text{B}=30^\circ+\text{x})$ $=\sin90^\circ$ $=1$ $=\text{R.H.S}$ $\text{Hence proved}.$

View full question & answer→

Question 82 Marks

If $\tan\text{A}=\frac{\text{5}}{\text{6}}$ and $\tan\text{B}=\frac{1}{\text{11}},$ then prove that $\text{A+B}=\frac{\pi}{4}$

Answer

We have,$\tan\text{A}=\frac{\text{5}}{\text{6}}$ and $\tan\text{B}=\frac{\text{1}}{\text{11}}$
Now, $\tan\text{(A+B)}=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}{\tan\text{B}}}$
$=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6}\times\frac{1}{11}}$
$=\frac{\frac{55+6}{66}}{1-\frac{5}{66}}$
$=\frac{\frac{61}{66}}{\frac{66-5}{66}}$
$=\frac{\frac{61}{66}}{\frac{61}{66}}$
$=\frac{61}{66}\times \frac{66}{ 61}$
$=1$
$\tan\frac{\pi}{4}$ $\Big[\because\tan\frac{\pi}{4}=1\Big]$
$\Rightarrow \tan\text{(A+B)}=\tan\frac{\pi}{4}$
$\Rightarrow\text{A+B}=\frac{\pi}{4}$
Hence proved.

View full question & answer→

Maths 2 Marks Questions

Take a timed test