1+ 1- =? - Vidyadip

MCQ

$\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}}=?$

A
$\text{cosec }\text{A}-\cot\text{A}$
✓
$\text{cosec }\text{A}+\cot\text{A}$
C
$\text{cosec }\text{A}\cot\text{A}$
D
None of these.

✓

Answer

Correct option: B.

$\text{cosec }\text{A}+\cot\text{A}$

$\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}}=\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}\times\frac{1+\cos\text{A}}{1+\cos\text{A}}}$
$=\sqrt{\frac{(1+\cos\text{A})^2}{1-\cos^2\text{A}}}$
$=\sqrt{\frac{(1+\cos\text{A})^2}{\sin^2\text{A}}}$
$=\frac{1+\cos\text{A}}{\sin\text{A}}$
$=\frac{1}{\sin\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}$
$=\text{cosec }\text{A}+\cot\text{A}$

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 "M.C.Q (1 Marks)" from Trigonometric Identities→Trigonometric Identities — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

Choose the correct alternative answer for each of the following question.
Which number cannot represent a probability?
A. 2/3
B. 1.5
C. 15%
D. 0.7

The zeros of the polynomial $4\text{x}^2+5\sqrt2\text{x}-3$ are:

The curved surface area of a cylinder is $264m^2$ and its volume is $924m^3$. The ratio of its diameter to its height is:

If $\frac{\text{x cosec}^230^\circ\sec^245^\circ}{8\cos^245^\circ\sin^260^\circ}=\tan^260^\circ-\tan^230^\circ,$ then $x =$

The material of a cone is converted into the shape of a cylinder of equal radius. If height of the cylinder is $5\ cm,$ then height of the cone is:

The standard form of quadratic equation $x-\frac{5}{x}=3 x-7$ is _____ .

A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is $h.$ If the total volume of the solid is $3$ times the volume of the cone, then the height of the circular is:

The solution of the equation 2x – y = 2 is ------.

Which one is the quadratic equation?

It is given that $\triangle\text{ABC}\sim\triangle\text{PQR}$ and $\frac{\text{BC}}{\text{QR}}=\frac{2}{3}$ then $\frac{\text{ar}(\triangle\text{PQR})}{\text{ar}(\triangle\text{ABC})}=?$