Question

If tan theta -cot theta =5 find (tan cube theta -cot cube theta )× (cot square theta -tan square theta)

Answer 1

Find the value:

Step-by-step explanation:

$\ Given \ value:\\\\\tan \theta - \cot \theta= 5\\\\\ find:\\\\(\tan^3 \theta -\cot^3 \theta) \times (\cot^2 \theta- \tan^2 \theta)\\\\\ Solution:\\\\\ formula:\\\\(a-b)^3 =a^3-b^3+3ab(a-b)\\\\\rightarrow (\tan \theta- \cot \theta)^3=\tan^3 \theta -\cot^3 \theta +3\tan \theta\times \cot \theta(\tan \theta- \cot \theta)\\\\\rightarrow (\tan \theta- \cot \theta)^3=\tan^3 \theta -\cot^3 \theta +3\tan \theta\times \frac{1}{\tan \theta}(\tan \theta- \cot \theta)$

$\rightarrow (\tan \theta- \cot \theta)^3=\tan^3 \theta -\cot^3 \theta +3(\tan \theta- \cot \theta)\\\\\therefore \tan \theta- \cot \theta= 5\\\\\rightarrow \tan^3 \theta -\cot^3 \theta= (\tan \theta- \cot \theta)^3+3(\tan \theta- \cot \theta)\\\\\rightarrow \tan^3 \theta -\cot^3 \theta= (5)^3+3(5)\\\\\rightarrow \tan^3 \theta -\cot^3 \theta= 125+15$

$\bold{\rightarrow \tan^3 \theta - \cot^3 \theta \ = 140}}$

$\cot^2 \theta -\tan^2 \theta= (\cot \theta - \tan \theta)(\cot \theta + \tan \theta)\\\\\cot^2 \theta -\tan^2 \theta= (-\cot \theta + \tan \theta)(\cot \theta + \tan \theta)\\\\\cot^2 \theta -\tan^2 \theta=- (\tan \theta -\cot \theta )(\cot \theta + \tan \theta)$

$\bold{\cot^2 \theta -\tan^2 \theta= -(5)(\sqrt{29})}$

$\ Equation:\\\\(\tan^3 \theta -\cot^3 \theta) \times (\cot^2 \theta- \tan^2 \theta)\\\\\rightarrow140 \times (-5\sqrt{29})\\\\\rightarrow -700\sqrt{29}$

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• Simplify: https://brainly.in/question/16255229

Answer 2

Answer:

Hope this will help you if this helps you put thanks

I don't no whether it is right if it is wrong sorry