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

Trigonometric Ratios of Multiples of an Angle

cosec A - 2co...

Question

cosec A - 2cot 2A cos A =

2sinA

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

secA

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

2cosAcotA

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

None of these

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

Open in App

Solution

The correct option is A
2sinA

cosec A - 2 cot 2A cos A = $\frac{1}{s i n A}$ - $\frac{2 c o s A c o s 2 A}{s i n 2 A}$

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

= 2sin A.

Suggest Corrections

Similar questions

\frac{cot A + c o s e c A - 1}{cot A - c o s e c A + 1} = \frac{1 + cos A}{cos A}

\frac{cos A - sin A + 1}{cos A + sin A - 1} = c o s e c A + cot A

c o s e c^{2} A = 1 + {cot}^{2} A

c o s e c A = \sqrt{2}

\frac{2 {sin}^{2} A + 3 {cot}^{2} A}{4 ({tan}^{2} A - {cos}^{2} A)}

A, 2 A

(i) cot A + tan A = \frac{1}{sin A cos A} = 2 cosec 2 A (i i) cot A - tan A = \frac{{cos}^{2} A - {sin}^{2} A}{sin A cos A} = 2 cot 2 A (i i i) 2 cot A = 2 (cosec 2 A + cot 2 A) \Rightarrow cosec 2 A + cot 2 A = cot A

f (x) + f (2 x) + f (4 x) + \dots + f (2^{n} x)

f (x)

g (x) - g (2 x)

f (x) + f (2 x) + f (4 x) + \dots + f (2^{n} x) = (g (x) - g (2 x)) + (g (2 x) - g (4 x)) + \dots (g (2^{n} x) - g (2^{n + 1} x)) = g (x) - g (2^{n + 1} x)

x

cosec 2 x + cosec 4 x + cosec 8 x + cosec 16 x + cosec 32 x

(1 + c o t A + t a n A) (s i n A - c o s A)

\frac{s e c A}{c o s e c^{2} A} - \frac{c o s e c A}{s e c^{2} A}

s i n A t a n A - c o t A c o s A

Join BYJU'S Learning Program