Question

please solve the problem TRIGONOMETRY​

Answer 1

Answer:

I am just 8th bro I don't about this and they haven't thought me this

Step-by-step explanation:

SORRY FOR THE INCONVENIENCE

Answer 2

Step-by-step explanation:

Given,

sec∅+tan∅ = m → (1)

W.K.T,

______________

Sec²∅-tan²∅ = 1

______________

=> (sec∅+tan∅)(sec∅-tan∅) = 1

[Since (a²-b²)=(a+b)(a-b) formula]

=> m(sec∅-tan∅) = 1

sec∅-tan∅ = 1/m → (2)

By solving equations (1) & (2),

Sec∅+tan∅ = m

Sec∅-tan∅ = 1/m

______________

2Sec∅ = m+(1/m)

2Sec∅ = (m²+1)/m

Sec∅ = [(m²+1)/m]/2

Sec∅ = (m²+1)/m × ½

:. Sec∅ = (m²+1)/2m

Substitute 'Sec∅' value in equation (1),

Sec∅+tan∅ = m

(m²+1)/2m + tan∅ = m

tan∅ = m - (m²+1)/2m

tan∅ = [2m²-(m²+1)]/2m

tan∅ = (2m²-m²-1)/2m

:. tan∅ = (m²-1)/2m

Now,

Tan∅ = sin∅/cos∅

Sin∅/Cos∅ = (m²-1)/2m

Sin∅.(1/Cos∅) = (m²-1)/2m

Sin∅.Sec∅ = (m²-1)/2m

Sin∅.(m²+1)/2m = (m²-1)/2m

Sin∅ = [(m²-1)/2m]/[(m²+1)2m]

Sin∅ = (m²-1)/2m × 2m/(m²+1)

Sin∅ = (m²-1)/(m²+1)

(Since both '2m' are cancel)

:. Sin∅ = (m²-1)/(m²+1).

I hope it helps you!!!