Answer:
refer to the pic
Step-by-step explanation:
hope it help...
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q(p2-1) = 2p
=>LHS = q(p2-1)
=>sec +cosec [(sin + cos)2-1
=> (sec+cosec) [sin2+cos2+2sin.cos -1]
= >(1/cos +1/sin) [1+2sin.cos-1]
{{identity: sin2=cos2=1}}
taking LCM of 1/sin + 1/cos
=> (sin +cos /sin.cos)[2sin.cos]
= >sin + cos / sin.cos * 2sin.cos
=> 2(sin + cos)
= >2(p) {{sin + cos = p
:(given in question)}}
Therefore 2p i.e. = RHS
hence, 2p = 2p
I HOPE ITS HELP YOU DEAR,
THANKS
Answer:
refer to the pic
Step-by-step explanation:
hope it help...