Sin 2A- A-1 2A- 2A-1 is equal to

The correct option is D A/2 Given expression is #10; 2sin Acos ^2A1 cos A #10;11 + cos 2A #10; #10; #10; = #160;2sin Acos ^2A1 co ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

sin 2A/ A-1 2...

Question

$(\frac{sin 2 A}{sec A - 1}) (\frac{sec 2 A}{sec 2 A + 1})$ is equal to

sin \frac{A}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cos \frac{A}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan \frac{A}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cot \frac{A}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $cot \frac{A}{2}$
Given expression is
$\frac{2 sin A {cos}^{2} A}{1 - cos A} \frac{1}{1 + cos 2 A}$

$= \frac{2 sin A {cos}^{2} A}{1 - cos A} \frac{1}{2 {cos}^{2} A}$

$= \frac{sin A}{1 - cos A}$

$= \frac{2 sin \frac{A}{2} cos \frac{A}{2}}{2 {sin}^{2} \frac{A}{2}}$
$= cot \frac{A}{2}$

Suggest Corrections

Similar questions

\frac{cos A}{1 - tan A} - \frac{{sin}^{2} A}{cos A - sin A} = sin A + cos A

Q. Prove that :

\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A} = \frac{2}{2 s i n^{2} A - 1} = \frac{2}{1 - 2 c o s^{2} A}

Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

Q. Solve

\frac{1 + sin A - cos A}{1 + sin A + cos A} = tan \frac{A}{2}

Q. (sina-cosa)^2=1-sin2a

Join BYJU'S Learning Program

(sin2AsecA−1)(sec2Asec2A+1) is equal to

The correct option is D cotA2Given expression is 2sinAcos2A1−cosA11+cos2A =2sinAcos2A1−cosA12cos2A =sinA1−cosA =2sinA2cosA22sin2A2 =cotA2

$(\frac{sin 2 A}{sec A - 1}) (\frac{sec 2 A}{sec 2 A + 1})$ is equal to