Question

Can anyone told me how to solve trigonometric problems myself

Answer 1

Answer:

You should remember:

Sin A=Perpendicular/Hypotenuse

Cosec A=1/Sin A

Cos A=Base/Hypotenuse

Sec A=1/Cos A

Tan A=Perpendicular/Base

Cot A=1/Tan A

Tan A=Sin A/Cos A

The value of trigonometric ratios for 0°,30°,45°,60°,90° should be remembered.

Sin(90-A)=Cos A and vice versa

Tan(90-A)=Cot A and vice versa

Sec(90-A)=CosecA and vice versa

Sin^2 A + Cos^2 A = 1

Sec^2 A = 1+Tan^2 A

Cot^2 A +1 = Cosec^2 A

Hope it helps you

Please mark as BRAINLIEST

Thank you

Answer 2

Answer:

ITS EASY

Step-by-step explanation:

JUST REMEMBER PANDIT BADRI PRASAD HAR HAR BOLE

P.B.P

H.H.B

SIN COS TAN (P/H, B/H , P/B )

A MANTRA TO REMEMBER

P means perpendicular

H means hypotenuse

B means base

Indentities

sin(θ) = 1/csc(θ)

cos(θ) = 1/sec(θ)

tan(θ) = 1/cot(θ)

And the other way around:

csc(θ) = 1/sin(θ)

sec(θ) = 1/cos(θ)

cot(θ) = 1/tan(θ)

And we also have:

cot(θ) = cos(θ)/sin(θ)