Question

tan cube theta by 1 + tan square theta + cos cube theta by 1 + cot square theta equal to sec theta × cosec theta - 2 Sin Theta into cos theta

Answer 1

Answer:here is ur answer.....

Step-by-step explanation:

May this help u....

Answer 2

To Prove:

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

Step-by-step explanation:

→ Formulas to be used:

(1.) 1 + tan²θ = sec²θ          (2.) 1 + cot²θ = cosec²θ         (3.) sin²θ + cos²θ = 1

→ Procedure:          

                              $L.H.S.=\frac{tan^{3}\theta }{1+tan^{2}\theta } +\frac{cot^{3}\theta}{1+cot^{2} \theta}$

                                          $=\frac{tan^{3}\theta}{sec^{2}\theta } +\frac{cot^{3}\theta }{cosec^{2} \theta}\\\\=tan^{3}\theta cos^{2}\theta+cot^{3}\theta sin^{2}\theta \\\\=(\frac{sin^{3} \theta}{cos^{3}\theta } )cos^{2}\theta+(\frac{cos^{3}\theta }{sin^{3}\theta } )sin^{2} \theta\\\\=\frac{sin^{3} \theta}{cos\theta }+\frac{cos^{3}\theta }{sin\theta } \\\\=\frac{sin^{4}\theta+cos^{4}\theta }{sin\theta cos\theta}$

→ Adding and subtracting 2sin²θcos²θ  in order to get (sin²θ + cos²θ)² :

                                          $=\frac{sin^{4}\theta +cos^{4}\theta +2sin^{2} \theta cos^{2} \theta-2sin^{2} \theta cos^{2} \theta}{sin\theta cos\theta}\\\\=\frac{(sin^{2}\theta +cos^{2}\theta)^{2} -2sin^{2} \theta cos^{2} \theta}{sin\theta cos\theta} \\\\=\frac{1 -2sin^{2} \theta cos^{2} \theta}{sin\theta cos\theta} \\\\=\frac{1}{sin\theta cos\theta}- \frac{2sin^{2} \theta cos^{2} \theta}{sin\theta cos\theta} \\\\=sec\theta cosec\theta-2sin\theta cos\theta=R.H.S.$

                                      ∴ L.H.S. = R.H.S. Proved

                     $\therefore\frac{tan^{3}\theta }{1+tan^{2}\theta } +\frac{cot^{3}\theta }{1+cot^{2} \theta} = sec\theta cosec\theta-2sin\theta cos\theta-Proved$

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