Math, asked by sandhyashanbhag63792, 1 year ago

If tan²A =1-a² then prove that secA + tan³AcosecA =(2-a²)³/2

Answers

Answered by jscott33
2

Answer:


Step-by-step explanation:

tan²θ=1-a²

LHS

=secθ+tan³θcosecθ

=√(1+tan²θ)+tan²θ×tanθ×√(1+cot²θ)

[∵, sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]

=√1+(1-a²)+(1-a²)×√(1-a²)×√{1+(1/tan²θ)}

=√(2-a²)+(1-a²)×√(1-a²)×√{1+1/(1-a²)}

=√(2-a²)+(1-a²)×√(1-a²)×√{(1-a²+1)/(1-a²)}

=√(2-a²)+(1-a²)×√(2-a²)

=√(2-a²)×(1+1-a²)

=√(2-a²)×(2-a²)

=(2-a²)¹/²⁺¹

=(2-a²)³/²

=RHS (Proved)





sandhyashanbhag63792: Can it be solved by the formula
sandhyashanbhag63792: n³/2 = root n²
Omkarsinghfzd: mtlb Kya hai iska
sandhyashanbhag63792: Formula
Answered by Omkarsinghfzd
1

it's answer and very short and easy to solve

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