Question

tan8a-tan5a-tan3a show that tan8a.tan5a.tan3a

Answer 1

$\textbf{Using}$

$\boxed{\bf\,tan(A+B)=\frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}}$

$tan\,8A=tan(5A+3A)$

$tan\,8A=\displaystyle\frac{tan\,5A+tan\,3A}{1-tan\,5A+tan\,3A}$

$\implies\,tan,8A[1-tan\,5A\,tan\,3A]=tan\,5A+tan\,3A$

$\implies\,tan8A-tan\,8A\,tan\,5A\,tan\,3A=tan\,5A+tan\,3A$

$\implies\boxed{\bf\,tan\,8A-tan\,5A-tan3A=tan\,8A\,tan\,5A\,tan\,3A}$

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Prove: (tanA+2)(2tanA+1)=5tanA+2sec^2 A

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Answer 2

Step-by-step explanation:

\textbf{Using}Using

\boxed{\bf\,tan(A+B)=\frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}}

tan(A+B)=

1−tanAtanB

tanA+tanB

tan\,8A=tan(5A+3A)tan8A=tan(5A+3A)

tan\,8A=\displaystyle\frac{tan\,5A+tan\,3A}{1-tan\,5A+tan\,3A}tan8A=

1−tan5A+tan3A

tan5A+tan3A

\implies\,tan,8A[1-tan\,5A\,tan\,3A]=tan\,5A+tan\,3A⟹tan,8A[1−tan5Atan3A]=tan5A+tan3A

\implies\,tan8A-tan\,8A\,tan\,5A\,tan\,3A=tan\,5A+tan\,3A⟹tan8A−tan8Atan5Atan3A=tan5A+tan3A

\implies\boxed{\bf\,tan\,8A-tan\,5A-tan3A=tan\,8A\,tan\,5A\,tan\,3A}⟹

tan8A−tan5A−tan3A=tan8Atan5Atan3A