Question

If \[\theta \] is an acute angle and \[\tan \theta +\cot \theta =2\] then \[{{\tan }^{7}}\theta +{{\cot }^{7}}\theta =2\]
A. True
B. False

Answer 1

Hint: We can solve this trigonometric equation \[\tan \theta +\cot \theta =2\] by using trigonometric formulas
\[\cot \theta =\dfrac{1}{\tan \theta }\] , \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] , \[{{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }\], \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] 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

Complete step-by-step answer:
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
So, the correct answer is “Option A”.

Note: Alternate approach ,We can also solve it by putting \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\] in given equation and on solving we get it as \[\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta }=2\] now we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] putting it and on cross multiply we get our equation as \[1=2\sin \theta \cos \theta \] and now the same process as above