tylermetzger
contestada

so my brother is seriously struggling with these every day and he is really sad right now it would mean a lot to me if you'd answer these 2 questions :)

1. csc2(x) - 2csc(x)cot(x) + cot2(x) = tan2 (x/2)

2. [cos cos (x) cos(y)] [tan tan (x) + tan tan(y)] = sin(x + y)

Respuesta :

iherhu

Answer:

here you go!

Step-by-step explanation:

Working with the left hand side

( 1 + cos x ) + ( 1 − cos x ) ( 1 − cos x ) ( 1 + cos x  1 + cos x + 1 − cos x 1 − cos 2 x  2 sin 2 x  2 csc 2 x

which is equal to the right hand side

By using  trigonometric identities both the questions are proved LHS = RHS. 1. csc2(x) - 2csc(x)cot(x) + cot2(x) = tan2 (x/2). 2. [cos (x) cos(y)] [tan tan (x) + tan(y)] = sin(x + y).

What are trigonometric identities?

Trigonometric identities are the functions that include trigonometric functions such as sine, cosine, tangents, secant, and, cot.

1. csc2(x) - 2csc(x)cot(x) + cot2(x) = tan2 (x/2)

Proof; LHS

[tex]csc^2(x) - 2csc(x)cot(x) + cot^2(x)\\\\(cscx - cot x)^2\\\\(\dfrac{1}{sin x} - \dfrac{cosx}{sin x} )^2\\\\\\(\dfrac{1-cosx}{sin x})^2\\ \\(tan)^2\dfrac{x}{2}\\[/tex]

Thus, LHS = RHS

2. [cos cos (x) cos(y)] [tan tan (x) + tan tan(y)] = sin(x + y)

Proof; LHS

[tex][cos (x) cos(y)] [tan (x) + tan(y)]\\\\ (cos (x) cos(y)) (\dfrac{sinx}{cos x} + \dfrac{siny}{cos y} )\\\\sin x \; cos y +sin y\; cos x\\\\sin (x+ y)[/tex]

Thus, LHS = RHS

Learn more about trigonometric;

https://brainly.com/question/21286835

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