Question

If tan² theta= (1-e^2), show that sec theta + tan theta cosec theta = (2-e^2)^2÷3​

Answer 1

Solution:

Given: tan²β = 1-e²

I think, your question is to prove: secβ+tan³β.cosecβ = $(2-e^{2})^{\frac{3}{2}}$

Taking L.H.S.,

= secβ+tan³β.cosecβ

= secβ+tan²β.tanβ.cosecβ

= $sec\beta+tan^{2}\beta.\frac{sin\beta}{cos\beta}.\frac{1}{sin\beta}$

= $sec\beta+tan^{2}\beta.\frac{1}{cos\beta}$

= secβ+tan²β.secβ

= secβ.(1+tan²β)

= ($\sqrt{1+tan^{2}\beta}$).(1+tan²β)

= $(1+tan^{2}\beta)^{\frac{1}{2}+1}$

= $(1+tan^{2}\beta)^{\frac{3}{2}}$

= $(1+1-e^{2})^{\frac{3}{2}}$

= $(2-e^{2})^{\frac{3}{2}}$ = R.H.S.

Hence Proved

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