Question

(tan^3theta÷1+ tan^3theta) + (cot^3theta÷1+cot^3theta) =
sectheta×cosectheta - 2 sintheta×costheta​

Answer 1

\frac{tan^{3}\theta}{1+tan^{2}\theta}+\frac{cot^{3}\theta}{1+cot^{2}\theta}\\= cosec\theta sec\theta - 2

Step-by-step explanation:

LHS = \frac{tan^{3}\theta}{1+tan^{2}\theta}+\frac{cot^{3}\theta}{1+cot^{2}\theta}\\=\frac{tan^{3}\theta}{sec^{2}\theta}+\frac{cot^{3}\theta}{cosec^{2}\theta}

\* We know the Trigonometric identities:

1. 1+tan²A = sec²A

2. 1+cot²A = cosec²A */

= \frac{\frac{sin^{3}\theta}{cos^{3}\theta}}{\frac{1}{cos^{2}\theta}}+\frac{\frac{cos^{3}\theta}{sin^{3}\theta}}{\frac{1}{sin^{2}\theta}}

After cancellation, we get

=\frac{sin^{3}\theta}{cos\theta}+\frac{cos^{3}\theta}{sin\theta}\\=\frac{(sin^{4}\theta+cos^{4}\theta)}{cos\theta sin\theta}\\=\frac{(sin^{2}\theta+cos^{2}\theta)^{2}-2sin\theta cos\theta}{cos\theta sin\theta}

/* We know that,

a²+b² = (a+b)²-2ab */

=\frac{(1-2sin\theta cos\theta)}{sin\theta cos\theta}

\* By Trigonometric identity:

sin²A + cos²A = 1 *\

= \frac{1}{sin\theta cos\theta} - \frac{2sin\theta cos\theta }{sin\theta cos\theta}\\= cosec\theta sec\theta - 2 \\= RHS

Therefore,

\frac{tan^{3}\theta}{1+tan^{2}\theta}+\frac{cot^{3}\theta}{1+cot^{2}\theta}\\= cosec\theta sec\theta - 2

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