Question

Tanx(cotx +tanx)= sec^2x

Answer 1

`\bf tan(x)[cot(x)+tan(x)]~~=~~ sec^2(x) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{sin(x)}{cos(x)}\left( \cfrac{cos(x)}{sin(x)}+ \cfrac{sin(x)}{cos(x)}\right)\implies \cfrac{sin(x)}{cos(x)}\left( \cfrac{cos^2(x)+sin^2(x)}{sin(x)cos(x)}\right) \\\\\\ \cfrac{\underline{sin(x)}}{cos(x)}\left( \cfrac{1}{\underline{sin(x)} cos(x)}\right)\implies \cfrac{1}{cos(x)cos(x)}\implies \cfrac{1}{cos^2(x)}\implies sec^2(x)`

recall sin²(x) + cos²(x) = 1.

Answer 2

Explanation:

Tanx(cotx +tanx)= sec^2x

distribute

tanx cot x + tan ^2 x = sec^2 x

tanx * cot x = sin/cos * cos /sin = 1

tanx cot x + tan ^2 x = sec^2 x

1 + tan ^2 x = sec ^2 x

1 + sin ^2 x/ cos ^2 x = 1 /cos ^2 x

get a common denominator

cos ^2 x/ cos ^2 x + sin ^2 x/cos ^2 x = 1 /cos ^2 x

combine the terms

(cos ^2 x + sin ^2 x) /cos ^2 x = 1 /cos ^2 x

sin ^2 x + cos ^ 2 x =1

1 /cos ^2 x = 1 /cos ^2 x