Question

Heya!! ❤️

Prove that :-

tan8α - tan5α - tan3α = tan8α tan5α tan3α


Thanks!​

Answer 1

$\underline{\textsf{Proof :}}$

$\mathsf{Now,\:tan8\alpha = tan(5\alpha+3\alpha)}$

$\to \mathsf{tan8\alpha=\frac{tan5\alpha+tan3\alpha}{1-tan5\alpha\:tan3\alpha}}$

$\to \mathsf{tan8\alpha(1-tan5\alpha\:tan3\alpha)}$
$\mathsf{=tan5\alpha+tan3\alpha}$

$\to \mathsf{tan8\alpha-tan8\alpha\:tan5\alpha\:tan3\alpha}$
$\mathsf{=tan5\alpha+tan3\alpha}$

$\to \boxed{\tiny{\mathsf{tan8\alpha-tan5\alpha-tan3\alpha=tan8\alpha\:tan5\alpha\:tan3\alpha}}}$

$\textsf{Hence, proved.}$

$\boxed{\underline{\textsf{More about Trigonometry :}}}$

$\:\:\:$It is the study of angles and the relationships between the angles and ratios sine, cosine, tangent, cosectant, sectant, cotangent.

$\underline{\textsf{Some formulas are -}}$

$\small{\textsf{1. sin(A + B) = sinA cosB + sinB cosA}}$

$\small{\textsf{2. cos(A + B) = cosA cosB - sinA sinB}}$

$\textsf{3. sin(A - B) = sinA cosB - sinB cosA}$

$\small{\textsf{4. cos(A - B) = cosA cosB + sinA sinB}}$

$\mathsf{5.\:tan(A + B) = \frac{tanA+tanB}{1-tanA\:tanB}}$

$\mathsf{6.\:tan(A - B) = \frac{tanA-tanB}{1+tanA\:tanB}}$

Answer 2

tan8A = tan(3A + 5A )

tan8A = tan3A + tan5A /1 - tan3A×tan5A

✓using formula tan(A+B) = tanA+tanB/1 - tanA×tanB

hence,

tan8A(1 - tan3A×tan5A) = tan3A + tan5A

tan8A - tan3A×tan5A×tan8A = tan3A+tan5A

tan3A - tan3A - tan5A =tan3A*tan5A*tan8A

[REARRANGING TERM]

Hence here Prooved

tan8A - tan3A - tan5A = tan3A*tan5A*tan8A