Explained: "Prove that cos⁡ec2A⋅sec2A=tan2A+cot2A+2"

Alex Cervantes

Answered question

2022-01-24

Prove that

\cos e c^{2} A \cdot \sec^{2} A = \tan^{2} A + \cot^{2} A + 2

Tapanuiwp

Beginner2022-01-25Added 13 answers

I assume your second

\csc^{2} A

was meant to be

\sec^{2} A

, viz.

\tan^{2} A + \cot^{2} A + 2 = {(\tan A + \cot A)}^{2} = {(\frac{\sin^{2} A + \cos^{2} A}{\sin A \cos A})}^{2}

= \frac{1}{\sin^{2} A \cos^{2} A} = \csc^{2} A \sec^{2} A

pripravyf

Beginner2022-01-26Added 12 answers

Let

s = \sin A and c = \cos A

. Then

\tan A = \frac{s}{c}

and

\cot A = \frac{c}{s}

. Therefore

R H S = \tan^{2} A + \cot^{2} A + 2 = {(\frac{s}{c})}^{2} + {(\frac{c}{s})}^{2} + 2

This gives

R H S = \frac{s^{4} + c^{4} + 2 s^{2} c^{2}}{s^{2} c^{2}} = \frac{{(s^{2} + c^{2})}^{2}}{s^{2} c^{2}}

But

s^{2} + c^{2} = \sin^{2} A + \cos^{2} A = 1

. Therefore

R H S = \frac{1}{s^{2} c^{2}} = {(\frac{1}{s})}^{2} {(\frac{1}{c})}^{2}

Since

\cos e c A = \frac{1}{s}

and

\sec A = \frac{1}{c}

, the claim follows (presumably you wanted to prove

\cos e c^{2} A \sec^{2} A = \tan^{2} A + \cot^{2} A + 2)

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