Question

plz help.............​

Answer 1

tanx + cotx = 2

tanx + cotx = 1 + 1

on comparing LHS and RHS

tanx = cotx = 1

${tan}^{7} x + {cot}^{7} x = {(1)}^{7} + {(1)}^{7} = 1 + 1 \\ = 2 \\ \\ hence \\ \\ {tan}^{7} x + {cot}^{7} x = 2$

Answer 2

Step-by-step explanation:

Hey user......

Trigo.......=_____=

Tan theta + Cot theta = 2

Tan theta+ 1/ Tan theta = 2

Tan^2 theta +1/ tan theta = 2

multiply by tan theta

Tan^2 theta + 1 = 2 tan theta

tan^2 theta - 2tan theta +1 =0

Let tan theta = x

x^2 -2x +1 =0

x = 1

Tan theta = 1

theta = π/4 = 45°

Now substituting in Problem equation!

(tan theta)^7 + (cot theta)^7

theta = 45° put it in equation

(tan 45°)^7 + (cot 45°)^7

1^7 + 1^7

= 2

So from this tremendous equation we get answer as :-

(tan theta)^7 + (cot theta)^7 = 2