Question

if p tan a=tan p.a so prove that ,sin²p.a/ sina=p²/1+(p²-1)sin a​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

By using properties of Ratios (Invertendo, Componendo), the problem can be solved with less effort.

We have

tanpA=ptan⁡A

∴sinpA/cospA=psinA/cosA

1. ∴sinpA/root(1-sin^2pA)=psinA/root (1-sin^2A

Squaring both sides, we obtain

sin^2pA/(1−sin^2pA)=p^2sin^2A/(1-sin^2A)

Applying Invertendo,

(1−sin^2pA)/sin^2pA=(1−sin^2A)/p^2 sin^2A

Applying Componendo,

(1−sin^2pA+sin^2pA)/sin^2pA=(1−sin^2A+p^2sin^2A)/p^2sin^2A

∴1/sin^2pA={1+(p2–1)sin^2A}/p^2sin^2A

∴sin^2A/sin^2pA={1+(p^2–1)}sin^2A/p^2

Finally, by applying Invertendo, we obtain the desired result:

sin^2pA/sin^2A=p^2/{1+(p2–1)sin^2A}

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