Question

tan + cot =6,find tan^2+ cot^2

Answer 1

$\tan( \alpha ) + \cot( \alpha ) = 6$
Squaring both sides
${( \tan( \alpha ) + \cot( \alpha ) ) }^{2} = {6}^{2} \\ { \tan }^{2} \alpha + { \cot}^{2} \alpha + 2 \tan( \alpha ) \cot( \alpha ) = 36$
tan×cot=1
Hence
${ \tan }^{2} \alpha + { \cot}^{2} \alpha = 36 - 2 \\ { \tan }^{2} \alpha + { \cot}^{2} \alpha = 34$

Answer 2

tan a + cot a = 6

take square both sides  

(tan a + cot a)² = 6²  

* Use formula: $( x + y )^{2}$ =  $x^2 + 2xy + y^2$

tan² a + 2 tan a . cot a + cot² a = 36  

we Know,  

tan a = $\frac{1}{ cot a }$

so,  

tan²  a + cot²  a + 2 = 36

tan²  a + cot²  a = 36 - 2

tan² a + cot² a = 34