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