Question

if tan θ+ cot θ=2,find the value of tan²θ+ cot² θ

Answer 1

${ tan \theta + cot \theta = 2 } \\ \\ \\ \bold{ \underline{Square \; on\; both\; sides ,}} \\ \\ \\ \\ ( tan \theta + cot \theta )^2 = 2^2 \\ \\ \\ tan^2 \theta + cot^2 \theta + 2( 1 ) = 4 \;\;\;\;\;\;\;\;\; | \bold{ by\; using ( a + b )^2 = a^2 + b^2 + 2ab \;\;\;\;\;\; and \;\;\;\;\; tan^2 \theta\; cot \theta = 1 } \\ \\ \\ \\ tan^2 \theta + cot^2 \theta + 2 = 4 \\ \\ \\ tan^2 \theta + cot^2 \theta = 4 -2 \\ \\ \\ tan^2 \theta + cot^2 \theta = 2 }$

Answer 2

Tan¢ + Cot¢ = 2


Squaring both sides , we get



( Tan¢ + Cot¢ )² = (2)²




Tan²¢ + Cot²¢ + 2Tan¢ Cot¢ = 4




Tan²¢ + Cot²¢ + 2 Tan¢ × 1/Tan¢ = 4



Tan²¢ + Cot²¢ + 2 × 1 = 4



Tan²¢ + cot²¢ = 4 - 2


Tan²¢ + Cot²¢ = 2