QUESTION IMAGE
Question
- write the expression tan x sin x in terms of cos x.
- write the expression \\( \frac { \tan ^ { 2 } x } { 1 + \tan ^ { 2 } x } \\) in terms of sin x.
- simplify \\( \frac { \cos x } { 1 - \sin x } - \tan x \\).
- simplify \\( \frac { 1 } { \sin ^ { 2 } x } - \frac { 1 } { \tan ^ { 2 } x } \\).
- simplify \\( \frac { \cos x } { \tan x } + \sin x \\).
- simplify \\( 1 + \frac { 1 } { \tan ^ { 2 } x } \\).
- simplify \\( \frac { 1 - \frac { 1 } { \cos ^ { 2 } x } } { 1 - \cos ^ { 2 } x } \\).
- prove that \\( \tan x + \frac { 1 } { \tan x } = \frac { 1 } { \sin x \cos x } \\).
Step1: Use the identity \(\tan x=\frac{\sin x}{\cos x}\)
Substitute \(\tan x=\frac{\sin x}{\cos x}\) into \(\tan x\sin x\).
We get \(\frac{\sin x}{\cos x}\cdot\sin x=\frac{\sin^{2}x}{\cos x}\).
Step2: Use the Pythagorean identity \(\sin^{2}x = 1-\cos^{2}x\)
Substitute \(\sin^{2}x = 1-\cos^{2}x\) into \(\frac{\sin^{2}x}{\cos x}\).
We have \(\frac{1 - \cos^{2}x}{\cos x}=\frac{1}{\cos x}-\cos x\).
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\(\frac{1}{\cos x}-\cos x\)