Question

Tan cube alpha by 1 + tan squared alpha + cos cube asha by 1 + cos theta is equal to feet and alpha cosec and alpha minus 2 sin alpha into cos alpha

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The proof is given below :

Step-by-step explanation:

= $\frac{tan^{3}\alpha }{1+tan^{2}\alpha } +\frac{cot^{3}\alpha }{1+cot^{2}\alpha }$

We know,

tan ∝ = sin ∝ / cos ∝

cot ∝ = cos ∝ / sin ∝

Substituting these values,

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

= $\frac{\frac{sin^{3}\alpha }{cos^{3}\alpha } }{\frac{sin^{2}\alpha+cos^{2}\alpha }{cos^{2}\alpha } }+\frac{\frac{cos^{3}\alpha }{sin^{3}\alpha } }{\frac{cos^{2}\alpha+sin^{2}\alpha }{sin^{2}\alpha } }$

We know,

sin²∝ + cos²∝ = 1

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

= $\frac{sin^{3}\alpha +cos^{2}\alpha}{cos^{3}\alpha} + \frac{cos^{3}\alpha +sin^{2}\alpha}{sin^{3}\alpha}$

= $\frac{sin^{3}\alpha}{cos\alpha} + \frac{cos^{3}\alpha}{sin\alpha}$

= $\frac{sin^{4}\alpha+cos^{4}\alpha}{sin\alpha+cos\alpha  }$

= $\frac{(sin^{2}\alpha )^{2}+(cos^{2}\alpha )^{2}}{sin\alpha cos\alpha  }$

= $\frac{(sin^{2}\alpha +cos^{2}\alpha)+2sin^{2}\alpha cos^{2}\alpha  }{sin\alpha cos\alpha }$

= $\frac{(1)+2sin^{2}\alpha cos^{2}\alpha  }{sin\alpha cos\alpha }$

= $\frac{1}{sin\alpha cos\alpha } +\frac{2sin^{2}\alpha cos^{2}\alpha}{sin\alpha cos\alpha}$

= sec∝ cosec∝ + 2sin∝cos∝

Hence Proved