Minor Trigonometric Identities

Tangent times Tangent Plus Cotangent

 * $$(2) \qquad \sec x-\cos x=\sin x\tan x$$

Proof: $$\sec x-\cos x=\frac{1-\cos^2x}{\cos x}=\frac{\sin^2x}{\cos x}=\sin x\tan x$$


 * $$(3) \qquad \tan^2x-\sin^2x=\tan^2x\sin^2x$$

Proof:  $$\tan^2x-\sin^2x=\frac{\sin^2x}{\cos^2x}-\frac{\sin^2x\cos^2x}{\cos^2x}=(1-\cos^2x)\frac{\sin^2x}{\cos^2x}=\sin^2x\tan^2x$$


 * $$(4) \qquad \sin^4x-\cos^4x=\sin^2x-cos^2x$$

Proof:  $$\sin^4x-\cos^4x=\sin^2x(1-\cos^2x)-\cos^2x(1-\sin^2x)=\sin^2x-\sin^2x\cos^2x-\cos^2x+\sin^2x\cos^2x=\sin^2x-\cos^2x$$


 * $$(5) \qquad \csc x-\sin x=\cos x\cot x$$

Proof: $$\csc x-\sin x=\frac{1-\sin^2x}{\sin x}=\frac{\cos^2x}{\sin x}=\cos x\cot x$$


 * $$(6) \qquad \sec^2x+\csc^2x=\sec^2x\csc^2x$$

Proof: $$\sec^2x+\csc^2x=\frac{\sin^2x+\cos^2x}{\cos^2x\sin^2x}=\frac{1}{\cos^2x\sin^2x}=\sec^2x\csc^2x$$


 * $$(7) \qquad \sec^4x-\tan^4x=\sec^2x+\tan^2x$$

Proof: $$\sec^4x-\tan^4x=\frac{1-\sin^4x}{\cos^4x}=\frac{1-\sin^2x(1-\cos^2x)}{\cos^2x(1-\sin^2x)}= \frac{1-\sin^2x+\sin^2x\cos^2x}{\cos^2x-\sin^2x\cos^2x}$$ $$ = \frac{\cos^2x+\sin^2x\cos^2x}{\cos^2x-\sin^2x\cos^2x}= \frac{1+\sin^2x}{1-\sin^2x}=\frac{1+\sin^2x}{\cos^2x}=\sec^2x+tan^2x$$


 * $$(8) \qquad \sin x\tan x+\cos x=\sec x$$

Proof: $$\sin x\tan x+\cos x=\frac{\sin^2x}{\cos x}+\cos x=\frac{\sin^2x+\cos^2x}{\cos x}=\frac{1}{\cos x}=\sec x$$


 * $$(9) \qquad \frac{1}{1-\sin x}+\frac{1}{1+\sin x}=2\sec^2x$$

Proof: $$\frac{1}{1-\sin x}+\frac{1}{1+\sin x}=\frac{1+\sin x+1-\sin x}{1-\sin^2x}=\frac{2}{\cos^2x}=2\sec^2x$$


 * $$(10) \qquad \frac{1}{1-\sin x}-\frac{1}{1+\sin x}=2\tan x\sec x$$

Proof: $$\frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{1+\sin x-1+\sin x}{1-\sin^2x}=\frac{2\sin x}{\cos^2x}=2\tan x\sec x$$


 * $$(11) \qquad \frac{\cos x}{\sec x+\tan x}=1-\sin x$$

Proof: $$\frac{\cos x}{\sec x+\tan x}=\frac{\cos^2x}{1+\sin x}=\frac{1-\sin^2x}{1+\sin x}=\frac{(1+\sin x)(1-\sin x)}{1+\sin x}=1-\sin x$$


 * $$(12) \qquad \frac{\sec x+1}{\sec^2x}=\frac{\sin^2x}{\sec x-1}$$

Proof: $$\frac{\sec x+1}{\sec^2x}=\cos^2x(\sec x+1)=\cos x+\cos^2x=\frac{1+\cos x}{\sec x}=\frac{1-\cos^2x}{\frac{1-\cos x}{\cos x}}=\frac{\sin^2x}{\sec x-1}$$


 * $$(13) \qquad (\sin x+\cos x)(\tan x+\cot x)=\sec x+\csc x$$

Proof: $$(\sin x+\cos x)(\tan x+\cot x)=(\sin x+\cos x)\frac{\sin^2x+\cos^2x}{\sin x\cos x}=\frac{\sin x+\cos x}{\sin x\cos x}=\sec x+\cos x$$


 * $$(14) \qquad \frac{\tan x}{\sec x+1}=\frac{\sec x-1}{\tan x}$$

Proof: $$\frac{\tan x}{\sec x+1}=\frac{\sin x}{1+\cos x}=\frac{\sin^2x}{\sin x(1+\cos x)}=\frac{1-\cos^2x}{\sin x(1+\cos x)}=\frac{1-\cos x}{\sin x}=\frac{\sec x-1}{\tan x}$$


 * $$(15) \qquad (a\cos x+b\sin x)^2+(b\cos x-a\sin x)^2=a^2+b^2$$

Proof: $$(a\cos x+b\sin x)^2+(b\cos x-a\sin x)^2=a^2\cos^2x+2ab\cos x\sin x+b^2\sin^2x+b^2\cos^2x-2ab\sin x\cos x+a^2\sin^2x$$ $$=a^2+b^2(\sin^2+\cos^2)=a^2+b^2$$


 * $$\frac{\sin^2x+2\cos x-1}{\sin^2x+3\cos x-3}=\frac{1}{1-\sec x}$$

Proof: $$\frac{\sin^2x+2\cos x-1}{\sin^2x+3\cos x-3}=\frac{1-\cos^2x-2\cos x-1}{1-\cos^2x+3\cos x-3}=\frac{\cos^2x-2\cos x}{\cos^2x-3\cos x +2}=\frac{\cos x(\cos x-2)}{(\cos x-1)(\cos x-2)}=\frac{\cos x}{\cos x-1}=\frac{1}{1-\sec x}$$


 * $$(16) \qquad \frac{\cos^2x+3\sin x-1}{\cos^2x+2\sin x+2}=\frac{1}{1+\csc x}$$

Proof: $$\frac{\cos^2x+3\sin x-1}{\cos^2x+2\sin x+2}=\frac{1-\sin^2x+3\sin x-1}{1-\sin^2x+2\sin x+2}=\frac{\sin^2x-3\sin x}{\sin^2x-2\sin x-3}=\frac{\sin x(\sin x-3)}{(\sin x-3)(\sin x+1)}=\frac{\sin x}{\sin x+1}=\frac{1}{1+\csc x}$$