Question

please help>

Answer 1

Solve the equation
cosec 2θ = sec θ + cot θ,
giving all solutions in the interval 0◦ < θ < 360◦

Answer 2

@ash2326 @Hero

Answer 3

Let's simplify it , I will use x instead of \(\theta\)
\[cosec\ 2x = \sec x +\cot x \]
\[\frac{1}{\sin 2x}=\frac{1}{\cos x }+\frac{\cos x }{\sin x}\]
\[\frac{1}{2\sin x \cos x }=\frac{\sin x +\cos ^2 x }{\cos x \sin x }\]
We can cancel sin x and cos x from both the sides, but we have to be mindful about one thing. We can only cancel if sin x and cos x are not 0, so answer can't have 180 degrees as part of the answer.

Do you get this part?

Answer 4

how did u come up with the first equation? what is the proof

Answer 5

\[cosec \ x =\frac 1 {\sin x}\]
\[\sec x = \frac 1 {\cos x }\]
\[\cot x =\frac {\cos x }{\sin x}\]
I just used these, which part you have doubt?

Answer 6

bro the question states that the equation in the RHS should be solved to form the the equation in the LHS

Answer 7

Where does it states that?

Solve the equation
cosec 2θ = sec θ + cot θ,
giving all solutions in the interval 0◦ < θ < 360◦

Answer 8

lol im sorry then i can do it...thanks alot bro:)

Answer 9

Cool :D