Question

solve it : cot2Φ.cot Φ = 1.

Answer 1

Given that - cot 2Φ.cotΦ = 1

solution - we can say that like this

cot 2Φ= 1/cot Φ = tan Φ

or. tanΦ = cot2Φ

or. = tan (π/2-2Φ)

tanΦ = tan [nπ + (π/2-2Φ)]

or. Φ = nπ + π/2-2Φ

or. Φ +2Φ = nπ+π/2

or. 3Φ = nπ + π/2

or. Φ = 1/3(nπ+π/2)

or. Φ = π/3(n+1/2)



I hope it's help you

Regards : Sangela

Answer 2

Your answer is ---

we have ,


cot2Φ.cotΦ = 1

=> cot2Φ = 1/cotΦ

=> cot2Φ = tanΦ [ °•° 1/cotΦ = tanΦ ]

=> tanΦ = cot2Φ

=> tanΦ = 1/2 (cotΦ - tanΦ) [ °•° cot2Φ = 1/2(cotΦ - tanΦ)

=> tanΦ = 1/2 (cotΦ - tanΦ)

=> 2tanΦ = cotΦ - tanΦ

=> 2tanΦ = 1/tanΦ - tanΦ

=> 2tanΦ = (1-tan^2Φ)/tanΦ

=> 2tan^2Φ = 1 - tan^2Φ

=> 3tan^2Φ = 1

=> tan^2Φ = 1/3

=> tanΦ = 1/√3

=> tanΦ = tan30° [ °•° tan30° = 1/√3 ]

=> Φ = 30°



【 Hope it helps you 】