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Dayanara Terry

Dayanara Terry

Answered question

2022-07-03

Prove 2 sec θ + 3 tan θ + 5 sin θ 7 cos θ + 5 2 tan θ + 3 sec θ + 5 cos θ + 7 sin θ + 8 = 1 cos θ sin θ

Answer & Explanation

isscacabby17

isscacabby17

Beginner2022-07-04Added 13 answers

Let cos θ = c , sin θ = s. Then,
s ( 2 + 3 s + 5 s c 7 c 2 + 5 c ) ( 1 c ) ( 2 s + 3 + 5 c 2 + 7 c s + 8 c ) = 5 c 3 + 5 c s 2 5 c + 3 c 2 + 3 s 2 3 = 5 c ( c 2 + s 2 1 ) + 3 ( c 2 + s 2 1 ) = 0
from which the claim follows.
Joel French

Joel French

Beginner2022-07-05Added 10 answers

hint 1 cos ( θ ) s i n ( θ ) = tan ( θ / 2 ) so now convert everything to tan by using sin ( 2 x ) = 2 t a n ( x ) 1 + tan 2 ( x ) , cos ( 2 x ) = 1 tan 2 ( x ) 1 + tan 2 ( x ) then just simplify it. as 1 + tan 2 ( x ) cancels off from the expression by taking lcm then use t a n ( x ) = t for algebraic simplifications. I dont think there's a nice short way to do it.

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