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A. True

B. False

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\[\cot \theta =\dfrac{1}{\tan \theta }\] , \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] , \[{{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }\], \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[Sin2\theta =2\sin \theta \cos \theta \]

Now we get value of \[\theta \] putting it in equation \[{{\tan }^{7}}\theta +{{\cot }^{7}}\theta =2\] and check whether it’s true or not

Given a Trigonometric equation \[\tan \theta +\cot \theta =2\] and we have to find whether equation \[{{\tan }^{7}}\theta +{{\cot }^{7}}\theta =2\] is true or false

Firstly, we can use formula \[\cot \theta =\dfrac{1}{\tan \theta }\] and write our equation as \[\tan \theta +\dfrac{1}{\tan \theta }=2\]

Now on solving our equation will look like as \[\dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2\], but we know that \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]

So, we can substitute \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] in this equation \[\dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2\]

On substituting we get our equation as \[\dfrac{{{\sec }^{2}}\theta }{\tan \theta }=2\]

Further it looks like \[{{\sec }^{2}}\theta =2\tan \theta \]

Now we know that \[{{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }\] and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]

So, we can substitute their values in this \[{{\sec }^{2}}\theta =2\tan \theta \] equation

Now our equation will look like \[\dfrac{1}{{{\cos }^{2}}\theta }=2\dfrac{\sin \theta }{\cos \theta }\]

On solving it as \[\dfrac{1}{\cos \theta }=2\dfrac{\sin \theta }{1}\] and Further on cross multiplying it looks like \[1=2\sin \theta \cos \theta \]

So finally, equation looks like \[1=2\sin \theta \cos \theta \] but we know that \[Sin2\theta =2\sin \theta \cos \theta \]

So, on substituting the value we get

It means \[\theta =(2n+1)\dfrac{\pi }{4}\] where n is a non-negative integer

But one condition for \[\theta \] is given that is \[\theta \] is an acute angle, which means \[0<\theta <\dfrac{\pi }{2}\]

So, it means \[n=0\] and \[\theta =\dfrac{\pi }{4}\]

Now we can put value \[\theta =\dfrac{\pi }{4}\] in \[{{\tan }^{7}}\theta +{{\cot }^{7}}\theta =2\]

On putting value of \[\theta \]equation will look like \[{{\tan }^{7}}\dfrac{\pi }{4}+{{\cot }^{7}}\dfrac{\pi }{4}=2\]

We know that \[\tan \dfrac{\pi }{4}={{\tan }^{7}}\dfrac{\pi }{4}=1\] and \[\cot \dfrac{\pi }{4}={{\cot }^{7}}\dfrac{\pi }{4}=1\]

So \[1+1=2\] LHS=RHS

Hence it is True statement