Question

prove the following identity : tan A + tan B whole divided by cot A + cot B = tan A . tan B​

Answer 1

Answer:

(tanA+ tan B)/cotA+.cotB

Step-by-step explanation:

(tanA+tanB)/(1/tanA+1/tanB

(tanA+tanB)/(tanA+tanB/tanA .tanB

(tanA+tanB)/(tanA .tanB /tanA +tanB

tanA .tan B

Answer 2

$\large\huge\red{\bold{\boxed{\boxed{\underline{\mathfrak{\green{\sf{QUESTION:-}}}}}}}}$

• prove the following identity : $\large{\mathrm{\large{\:\frac{(\tan A +\tan B)}{(\cot A + \cot B)}}\: =\: \tan A \times\tan B}}$

$\large\huge{\boxed{\boxed{\underline{\mathfrak{\green{\sf{PROVE:-}}}}}}}$

Take L.H.S.

$\large{\hookrightarrow{\mathrm{\:\frac{(\tan A +\tan B)}{(\cot A + \cot B)}}}}$

$\large{\hookrightarrow{\mathrm{\:\frac{\frac{\sin A}{\cos A}\:+\frac{\sin B}{\cos B}}{\frac{\cos A}{\sin A}\:+\frac{\cos B}{\sin B}}}}}$

$\large{\hookrightarrow{\mathrm{\:\frac{\frac{\cancel{(\sin A \cos B)+(\sin B \cos A)}}{(\cos A \cos B)}}{\frac{\cancel{(\sin A \cos B)+(\sin B \cos A)}}{(\sin A \sin B)}}}}}$

$\large{\hookrightarrow{\mathrm{\:\frac{(\sin A \sin B)}{(\cos A \cos B)}}}}$

$\large{\hookrightarrow{\mathrm{\:\frac{\sin A}{\cos A}\times\frac{\sin B}{\cos B}}}}$

$\bold{\pink{\boxed{\boxed{\mathrm{(\:\tan A \tan B)}}}}}$

Proved

$\large{\underline{\underline{\mathfrak{\green{\sf{USING\:FORMULA:-}}}}}}$.

❏ $\red{\mathrm{\tan X \:=\frac{\sin X}{\cos X}}}$

❏ $\red{\mathrm{\cot X\:=\frac{\cos X}{\sin X}}}$

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