tan 43=(cos 2-sin 2)/(cos 2+sin 2)
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LHS=RHS
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Savit11:
ok
please
tan2°=sin2°/cos2°
then we can multiply cos2° with the upper and lower part
how you write 1-tan 2 as cos 2
o i understood
ok batati hu
aap konse class me hay?
ye batao pahle
thanks a lot to you
Answered by
5
Answer:
Consider a right-angled triangle, ΔABC, where ∠BAC=θ,

By the Pythagorean theorem,AC2+BC2=AB2Dividing by AB2,⇒AC2AB2+BC2AB2=AB2AB2⇒(oppositehypotenuse)2+(adjacenthypotenuse)2=AB2AB2=1⇒sin2θ+cos2θ=1
Step-by-step explanation:
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