How do you prove that $\frac{{sec}^{2} θ}{2} = (2 - 2 {cos}^{2} θ) {csc}^{2} (2 θ)$ $(sec^2theta)/2 = (2-2cos^2theta)csc^2(2theta)$ ?

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How do you prove that $\frac{{sec}^{2} θ}{2} = (2 - 2 {cos}^{2} θ) {csc}^{2} (2 θ)$ $(sec^2theta)/2 = (2-2cos^2theta)csc^2(2theta)$ ?

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Answer 1 · 2016-02-26T08:49:30

see below

Explanation:

RHS = $2 (1 - {cos}^{2} θ) \cdot \frac{1}{{sin}^{2} (2 θ)}$ $2(1-cos^2theta)*1/sin^2(2theta)$
= $2 (1 - {cos}^{2} θ) \cdot \frac{1}{4 {sin}^{2} θ \cdot {cos}^{2} θ}$ $2(1-cos^2theta)*1/(4sin^2theta*cos^2theta)$
= $2 ({sin}^{2} θ) \cdot \frac{1}{4 {sin}^{2} θ \cdot {cos}^{2} θ}$ $2(sin^2theta)*1/(4sin^2theta*cos^2theta)$
= $\frac{{sec}^{2} θ}{2}$ $sec^2theta/2$ = LHS

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How do you prove that $\frac{{sec}^{2} θ}{2} = (2 - 2 {cos}^{2} θ) {csc}^{2} (2 θ)$ $(sec^2theta)/2 = (2-2cos^2theta)csc^2(2theta)$ ?

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Explanation:

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How do you prove that sec2θ2=(2−2cos2θ)csc2(2θ) (sec^2theta)/2 = (2-2cos^2theta)csc^2(2theta) ?

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How do you prove that $\frac{{sec}^{2} θ}{2} = (2 - 2 {cos}^{2} θ) {csc}^{2} (2 θ)$ $(sec^2theta)/2 = (2-2cos^2theta)csc^2(2theta)$ ?