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Sav [38]
3 years ago
7

Express all trigonometric ratio in theform of sintita​

Mathematics
2 answers:
Gennadij [26K]3 years ago
8 0

Answer:

Sin theta = Perpendicular/Hypotenuse = a/√(a^2+b^2)

cos theta = 1- sintheta = 1- a/√(a^2+b^2)

Tan theta = sin/cos = tan = sin/1-sin

= a/√(a^2+b^2)/√(a^2+b^2)-a/(a^2+b^2)

= a/√(a^2+b^2) × (a^2+b^2)/(a^2+b^2)-a

= a/(a^2+b^2)-a

So cosec theta = inverse of sine theta = (a^2+b^2)/a

Sec theta = inverse of cos theta = a-(a^2+b^2)/a

Cotan theta = inverse of tan theta = (a^2+b^2)-a/a

Hope it helps

bonufazy [111]3 years ago
6 0

Answer:

there you go :-  

cos a = square root of (1- square of sin a)  

tan = sin a / cos a = sin a/(square root of (1- square of sin a))  

cosec a = 1/sin a  

sec a = 1/cos a = 1/(square root of (1- square of sin a))  

cot a = cos a /sin a = (square root of (1- square of sin a))/sin a

Step-by-step explanation:

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