sin theta is equal to a square a square minus b square upon a square + b square find the value of tan theta and sec theta a square minus b square upon 2 sb ,a square + b square upon 2ab
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Step-by-step explanation:
We know that,
Sin ¢ = p/h Cosec ¢ = h/p
Cos ¢ = b/h Sec ¢ = h/b
tan ¢ = p/b Cot ¢ = b/p
Here P = perpendicular, b= base and h = hypotenuse.
Given, Sin ¢ = (a² - b²) / (a² + b²)
here, perpendicular (p) = (a² - b²) and hypotenuse (h) = (a² + b²)
by, Pythagoras theorem.
base (b) = √(a² - b²)² + (a² - b²)²
base (b) = 2ab
Now...
Cos ¢ = (2ab) / (a² + b²)
tan ¢ = (a² - b²) / 2ab
Cosec ¢ = (a² + b²)/ (a² - b²)
Sec ¢ = (a² + b²)/ 2ab
cot ¢ = 2ab / (a² - b²)
Step-by-step explanation:
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