Download App

Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
If $ \tan\text{A}+\tan\text{B}=\text{a}$ and $\cot\text{A}+\cot\text{B}=\text{b},$ prove that $\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}.$
✓

Answer

We 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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Trigonometric Functions→Trigonometric Functions — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find the value of (1.01)10 + (1 - 0.01)10 correct to 7 places of decimal.

→
Evaluate the following:

$\sum_\limits{\text{k}=1}^{\text{n}}(2^\text{k}+3^{\text{k}-1})$

→
Three vertices of a Parallelogram ABCD are A(1, 2, 3), B(-1, -2, -1) and C(2, 3, 2). Find the fourth vertex D.
→
$\text{x}^4+4\text{x}^3+6\text{x}^2+4\text{x}+9,$ when $\text{x}=-1+\text{i}\sqrt{2}$
→
Find a point on the x-axis, which is equidistant from the point (7, 6) and (3, 4).
→
Find the equation of the ellipse in the following case:
eccentricity $\text{e}=\frac{1}{2}$ and foci $(\pm2, 0)$
→
$\cot\text{x}+\cot\Big(\frac{\pi}{3}-\text{x}\Big)+\cot\Big(\frac{\pi}{3}-\text{x}\Big)=3\cot3\text{x}$
→
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
→
Prove that: $\frac{\sin\text{(A-B)}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{(B-C)}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{(C-A)}}{\sin\text{C}\sin\text{A}}=0$
→
Find the equations to the diagonals of the rectangle the equations of whose sides are x = a, x = a', y = b and y = b'.
→