Evalute tan⁡α−cot⁡αsin4α−cos4α if α is an acute angle and tan⁡α=2

Hints: tan⁡a=1cot⁡a cos2⁡a=11+tan2⁡a and sin2⁡a=11+cot2⁡a

Factorizing the denominator and using the fact that sin2⁡a+cos2⁡a=1, we have tan⁡α−cot⁡αsin4⁡α−cos4⁡α=tan⁡α−cot⁡αsin2⁡α−cos2⁡α=sec2⁡αtan⁡α−cot⁡αtan2⁡α−1=(tan2⁡α+1)tan⁡α−tan⁡αtan2⁡α−1 =(4+1)(2−12)4−1=52

Hints: tan⁡a=1cot⁡a cos2⁡a=11+tan2⁡a and sin2⁡a=11+cot2⁡a

Factorizing the denominator and using the fact that sin2⁡a+cos2⁡a=1, we have tan⁡α−cot⁡αsin4⁡α−cos4⁡α=tan⁡α−cot⁡αsin2⁡α−cos2⁡α=sec2⁡αtan⁡α−cot⁡αtan2⁡α−1=(tan2⁡α+1)tan⁡α−tan⁡αtan2⁡α−1 =(4+1)(2−12)4−1=52

Get the solution to "Evalute tan⁡α−cot⁡αsin4α−cos4α if α is an acute angle..."

feminizmiki

Answered question

2022-01-16

Evalute

\frac{\tan α - \cot α}{\sin^{4} α - \cos^{4} α}

α

is an acute angle and

\tan α = 2

Answer & Explanation

RizerMix

Expert2022-01-20Added 656 answers

Hints:

\tan a = \frac{1}{\cot a}

\cos^{2} a = \frac{1}{1 + \tan^{2} a} and \sin^{2} a = \frac{1}{1 + \cot^{2} a}

Vasquez

Expert2022-01-20Added 669 answers

With openspaces

user_27qwe

Skilled2022-01-23Added 375 answers

Factorizing the denominator and using the fact that

\sin^{2} a + \cos^{2} a = 1

, we have

\frac{\tan α - \cot α}{\sin^{4} α - \cos^{4} α} = \frac{\tan α - \cot α}{\sin^{2} α - \cos^{2} α} = \sec^{2} α \frac{\tan α - \cot α}{\tan^{2} α - 1} = (\tan^{2} α + 1) \frac{\tan α - \tan α}{\tan^{2} α - 1}

= (4 + 1) \frac{(2 - \frac{1}{2})}{4 - 1} = \frac{5}{2}

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