Question

If (tan theta + cot theta ) = 7/4 then (tan^2 theta +cot^2 theta) = ?​

Answer 1

Answer:

this is yr answer ..follow me

Answer 2

If (tan theta + cot theta ) = 7/4 then (tan^2 theta +cot^2 theta) = $\dfrac{17}{16}$

Step-by-step explanation:

Given $tan\theta +cot\theta =\dfrac{7}{4}$

We have to find $tan^{2}\theta +cot^{2}\theta$

We know the identity $\left ( \textrm{a+b} \right)^{2}=\textrm{a}^{2}+\textrm{b}^{2}+2\textrm{ab}$

So we have

$\left (tan\theta +cot\theta \right )^{2}=tan^{2}\theta +cot^{2} \theta +2tan\theta cot\theta$

$\Rightarrow tan^{2}\theta +cot^{2} \theta =\left (tan\theta +cot\theta \right )^{2}-2tan\theta cot\theta$

$\Rightarrow tan^{2}\theta +cot^{2} \theta =\left (\dfrac{7}{4} \right )^{2}-2tan\theta \times\frac{1}{tan\theta }$

$\Rightarrow tan^{2}\theta +cot^{2} \theta =\left (\dfrac{7}{4} \right )^{2}-2$

$\Rightarrow tan^{2}\theta +cot^{2} \theta =\dfrac{49}{16}-2$

$\Rightarrow tan^{2}\theta +cot^{2} \theta =\dfrac{49-32}{16}$

$\Rightarrow tan^{2}\theta +cot^{2} \theta =\dfrac{17}{16}$