MCQ
$\text{If }\cos\text{A}=\text{m}\cos\text{B},\text{ than }\cot\frac{\text{A+B}}{2}\cot\frac{\text{B}-\text{A}}{2}=$
  • A
    $\frac{\text{m}-1}{\text{m}+1}$
  • B
    $\frac{\text{m}+2}{\text{m}-2}$
  • $\frac{\text{m}+1}{\text{m-1}}$
  • D
    None of these

Answer

Correct option: C.
$\frac{\text{m}+1}{\text{m-1}}$
Given:
$\cos\text{A}=\text{m}\cos\text{B}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}=\frac{\text{m}}{1}$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{A}-\cos\text{B}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)}{-2\sin\Big(\frac{\text{B+A}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)\\\text{ and } \cos\text{A}-\cos\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$\Rightarrow\ \frac{\cos\Big(\frac{\text{B}-\text{A}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}{\sin\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{B}-\text{A}}{2}\Big)=\frac{\text{m}+1}{\text{m}-1}$

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

Similar questions

The line $4x -3y + 15 = 0$ intersect the circle $x^2 + y^2 -6x -8y = 0$ at two points $A$ and $B$. Maximum area of $\Delta ABC$, where $C$ is a point on circumference of circle, will be - .............. $\mathrm{sq.\, units}$
if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to
The value of $9^{\frac13}.9^{\frac19}.9^{\frac{1}{27}}\dots\text{to }\infty,$ is:
Which of the following is a set?
A. A collection of vowels in English alphabets is a set.
B. The collection of most talented writers of India is a set.
C. The collection of most difficult topics in Mathematics is a set.
D. The collection of good cricket players of India is a set.
The sides of a triangle are $\sin \alpha ,\,\cos \alpha $ and $\sqrt {1 + \sin \alpha \cos \alpha } $ for some $0 < \alpha < \frac{\pi }{2}$. Then the greatest angle of the triangle is.....$^o$
$\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{{\left( {1 + \left\{ x \right\}} \right)}^{\frac{1}{{\left\{ x \right\}}}}} - \frac{e}{{\sqrt {{e^{\left\{ x \right\}}}} }}}}{{1 - \cos \left\{ x \right\}}}$ (where {.} denotes fractional part function)
If $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ are the eccentricities of the ellipse, $\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1$ and the hyperbola, $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$ respectively and $\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ is a point on the ellipse, $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},$ then $\mathrm{k}$ is equal to 
Choose the correct answer.
The real value of $\alpha$ for which the expression $\frac{1-\text{i}\sin\alpha}{1+2\text{i}\sin\alpha}$ is purely real is:
The area of an equilateral triangle inscribed in the circle $x^2+ y^2 - 6x - 8y - 25 = 0$ is:
A line passing through origin and is perpendicular to two given lines $2x + y + 6 = 0$ and $4x + 2y - 9 = 0$, then the ratio in which the origin divides this line is