Question

Prove that cot A + tan ( π + A ) + ( π/2 + A ) + tan 2π - A ) = 0​

Answer 1

$\cot A$ + $\tan ( \pi+ A )$ + $\tan (\dfrac{\pi}{2} +A)$ + $\tan (2\pi - A)$ = 0​, proved.

Step-by-step explanation:

To prove that: $\cot A$ + $\tan ( \pi+ A )$ + $\tan (\dfrac{\pi}{2} +A)$ + $\tan (2\pi - A)$ = 0​.

L.H.S. = $\cot A$ + $\tan ( \pi+ A )$ + $\tan (\dfrac{\pi}{2} +A)$ + $\tan (2\pi - A)$

Using the trigonometric identities:

$\tan ( \pi+ A )$ = $\tan A$

$\tan (\dfrac{\pi}{2} +A)$ = - $\cot A$ and

$\tan (2\pi - A)$ = - $\tan A$

= $\cot A$ + $\tan A$  + (- $\cot A$) + (- $\tan A$)

= $\cot A$ + $\tan A$ - $\cot A$ - $\tan A$

= ($\cot A$ - $\cot A$) + ($\tan A$ - $\tan A$)

= 0 + 0

= 0

= R.H.S., proved.

∴ $\cot A$ + $\tan ( \pi+ A )$ + $\tan (\dfrac{\pi}{2} +A)$ + $\tan (2\pi - A)$ = 0​, proved.

Answer 2

Step-by-step explanation:

Consider LHS:-

( π = 180 )

cot A + tan (π + A) + tan ( π / 2 + A) + tan (2π - A)

cot A + tan (180+A)+tan (180/2 +A)+ tan(2*180-A)

cot A +tan (180+A)+tan ( 90 + A) + tan (360 - A)

tan (180 + A) = tan A

tan (90 + A) = -cot A

tan (360 - A)= -tan A

cot A + tan A - cot A - tan A = 0 = RHS

Hence proved..