Question

tan5A - tan3A÷tan 5A+Tan3A=sin2A÷sin8A
prove that

Answer 1

Answer:

Formula used:

sin(A+B) = sinA cosB + cosA sinB

sin(A-B) = sinA cosB - cosA sinB

\frac{tan5A-tan3A}{tan5A+tan3A}

=\frac{\frac{sin5A}{cos5A}-\frac{sin3A}{cos3A}}{\frac{sin5A}{cos5A}+\frac{sin3A}{cos3A}}

=\frac{sin5A.cos3A-cos5A.sin3A}{sin5A.cos3A+cos5A.sin3A}

=\frac{sin(5A-3A)}{sin(5A+3A)}

=\frac{sin2A}{sin8A}

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hope this helps u..

Answer 2

Answer:

Tan2A÷tan8A

(Sin2A÷cos2A)÷(sin8A÷Cos8A)

÷by cosA upward and downward

So finally we get

Sin2A ÷sin8A