Question

Prove that tan x + cot x = sec x . Cose x.

Answer 1

Dear student,


$$\begin{gathered} \tan x+cotx \\ =\frac{\sin x}{\cos x}+\frac{\cos {\displaystyle }{\displaystyle x}}{\sin {\displaystyle }{\displaystyle x}} \\ =\frac{{\displaystyle {\sin }^{2}x+{\cos }^{2}x}}{\cos {\displaystyle }{\displaystyle x}{\displaystyle }{\displaystyle .}{\displaystyle }{\displaystyle \sin }{\displaystyle }{\displaystyle x}}\{{\sin }^{2}x+{\cos }^{2}x=1\} \\ =\frac{1}{\cos {\displaystyle }{\displaystyle x}{\displaystyle .}{\displaystyle }{\displaystyle \sin }{\displaystyle }{\displaystyle x}}\{\frac{1}{\cos {\displaystyle }{\displaystyle x}}=secx\&\frac{1}{{\displaystyle \sin x}}=\cos ecx\} \\ =secx.\cos ecx \end{gathered}$$

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