Question

please solve my question correcty.......don't spam......and explanation must....​

Answer 1

Step-by-step explanation:

To prove: cot(2θ) + tan(θ)=cosec(2θ)

Proof:

Let us put θ=α in the given question.

so,

$\cot(2 \alpha ) + \tan( \alpha ) = \frac{1}{ \tan(2 \alpha ) } + \tan( \alpha )$

$= \frac{1}{ \frac{2 \tan( \alpha ) }{1 - \tan^{2} ( \alpha ) } } + \tan( \alpha )$

$= \frac{1 - \tan^{2} ( \alpha ) }{2 \tan( \alpha ) } + \tan( \alpha )$

Taking lcm,

$\frac{1 - \tan^{2} ( \alpha ) + 2 \tan^{2} ( \alpha ) }{2 \tan( \alpha ) }$

$= \frac{1 + \tan^{2} ( \alpha ) }{2 \tan( \alpha ) }$

$= \frac{1}{ \frac{2 \tan( \alpha ) }{1 + \tan^{2} ( \alpha ) } }$

$= \frac{1}{ \sin(2 \alpha ) }$

$= \csc(2 \alpha )$

So, cot(2θ) + tan(θ)= cosec(2θ), (putting back θ=α)

Here formula used are .......

tan(2θ)={2tan(θ)}/{1-tan²(θ)} and

sin(2θ)={2tan(θ)}/{1+tan²(θ)}

Answer 2

Answer:

cos(a-b)=cosa.cosb+sina.sinb

tana =sina/cosa

cota = cosa/sina

cot2a + tana

=(cos2a/sin2a)+(sina/ cosa)

take LCM

=(cos2a.cosa+sinasin2a)/sin2a.cosa

by formula,

=cosa/sin2a.cosa

=1/sin2a

ie.,cosec2a . (: a instead of theta)