Question

If theta is an acute angle and tan theta + cot theta equal to 2 then the value of sin cube
theta + cos cube theta is

Answer 1

Answer:

The answer is $\dfrac{1}{\sqrt 2}$

Step-by-step explanation:

Given

     $\tan\theta+\cot\theta=2$

Using $\tan\theta\cot\theta=1$

    $\Rightarrow \tan\theta+\dfrac{1}{\tan\theta}=2\\\Rightarrow \tan^2\theta+1=2\tan\theta\\\Rightarrow \tan^2\theta-2\tan\theta+1=0\\\Rightarrow (\tan\theta-1)^2=0\\\Rightarrow \tan\theta=1\Rightarrow \theta=45^\circ$

Therefore,

     $\sin^3\theta+\cos^3\theta=(\sin 45^\circ)^3+(\cos 45^\circ)^3$

                            $=\frac{1}{(\sqrt 2)^3}+\frac{1}{(\sqrt 2)^3}=2(\frac{1}{(\sqrt 2)^3})\\\\=2(\frac{1}{(\sqrt 2)^2})(\frac{1}{\sqrt 2})=2(\frac{1}{2}(\frac{1}{\sqrt 2})=\dfrac{1}{\sqrt 2}$

Answer 2

Answer:

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