Trigonometry Identities (Exercise 8.3, Class 10)

This lesson works through seven identity proofs from Exercise 8.3 on trigonometry. In every proof we begin with the left side, rewrite the trigonometric ratios in terms of $\sin$ and $\cos$, and simplify until we reach the right side. Remember the key relations $\sec\theta=\tfrac{1}{\cos\theta}$, $\csc\theta=\tfrac{1}{\sin\theta}$, $\tan\theta=\tfrac{\sin\theta}{\cos\theta}$, $\cot\theta=\tfrac{\cos\theta}{\sin\theta}$, and the Pythagorean identities.

Prove $\sec A\,(1-\sin A)(\sec A+\tan A)=1$

Start from the left side and write everything over $\cos A$:

$\sec A\,(1-\sin A)(\sec A+\tan A)=\frac{1}{\cos A}\,(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right).$

The second bracket has a common denominator, so add the numerators:

$=\frac{1}{\cos A}\,(1-\sin A)\cdot\frac{1+\sin A}{\cos A}=\frac{(1-\sin A)(1+\sin A)}{\cos^2 A}.$

The numerator is a difference of squares, and $1-\sin^2 A=\cos^2 A$:

$=\frac{1-\sin^2 A}{\cos^2 A}=\frac{\cos^2 A}{\cos^2 A}=1.$

Prove $9\sec^2 A-9\tan^2 A=9$

Take the common factor $9$ out of the left side:

$9\sec^2 A-9\tan^2 A=9(\sec^2 A-\tan^2 A).$

Since $\sec^2 A-\tan^2 A=1$, this equals $9\cdot 1=9$.

Prove $(\sec A+\tan A)(1-\sin A)=\cos A$

Rewrite the first bracket over $\cos A$:

$(\sec A+\tan A)(1-\sin A)=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)=\frac{1+\sin A}{\cos A}\,(1-\sin A).$

Multiply the numerators and use the difference of squares:

$=\frac{(1+\sin A)(1-\sin A)}{\cos A}=\frac{1-\sin^2 A}{\cos A}=\frac{\cos^2 A}{\cos A}=\cos A.$

Prove $\dfrac{1+\tan^2 A}{1+\cot^2 A}=\tan^2 A$

Use the identities $1+\tan^2 A=\sec^2 A$ and $1+\cot^2 A=\csc^2 A$:

$\frac{1+\tan^2 A}{1+\cot^2 A}=\frac{\sec^2 A}{\csc^2 A}.$

Now write each in terms of $\sin$ and $\cos$, where $\sec^2 A=\tfrac{1}{\cos^2 A}$ and $\csc^2 A=\tfrac{1}{\sin^2 A}$:

$=\frac{\tfrac{1}{\cos^2 A}}{\tfrac{1}{\sin^2 A}}=\frac{\sin^2 A}{\cos^2 A}=\tan^2 A.$

Prove $(\csc\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\theta}$

Write the left side in terms of $\sin$ and $\cos$:

$(\csc\theta-\cot\theta)^2=\left(\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\right)^2=\frac{(1-\cos\theta)^2}{\sin^2\theta}.$

Replace $\sin^2\theta=1-\cos^2\theta=(1+\cos\theta)(1-\cos\theta)$:

$=\frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}.$

Cancel one factor of $1-\cos\theta$:

$=\frac{1-\cos\theta}{1+\cos\theta}.$

Prove $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Take the common denominator $(1+\sin A)\cos A$:

$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=\frac{\cos^2 A+(1+\sin A)^2}{(1+\sin A)\cos A}.$

Expand $(1+\sin A)^2=1+2\sin A+\sin^2 A$:

$=\frac{\cos^2 A+1+2\sin A+\sin^2 A}{(1+\sin A)\cos A}.$

Since $\cos^2 A+\sin^2 A=1$, the numerator becomes $2+2\sin A=2(1+\sin A)$:

$=\frac{2(1+\sin A)}{(1+\sin A)\cos A}=\frac{2}{\cos A}=2\sec A.$

Prove $(1+\tan\theta+\sec\theta)(1+\cot\theta-\csc\theta)=2$

Write each bracket in terms of $\sin$ and $\cos$:

$\left(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\right)\left(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\right)=\frac{\cos\theta+\sin\theta+1}{\cos\theta}\cdot\frac{\sin\theta+\cos\theta-1}{\sin\theta}.$

Group $\sin\theta+\cos\theta$ as one term, so the numerator is of the form $(a+1)(a-1)$ with $a=\sin\theta+\cos\theta$:

$=\frac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta}.$

Expand $(\sin\theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta$:

$=\frac{1+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2.$

Key takeaways

• To prove an identity, transform one side step by step; converting every ratio to $\sin$ and $\cos$ almost always opens up the simplification.
• The Pythagorean identities $\sin^2\theta+\cos^2\theta=1$, $\sec^2\theta-\tan^2\theta=1$, and $1+\cot^2\theta=\csc^2\theta$ are the main tools for replacing and cancelling.
• Difference of squares and common factors let you cancel matching pieces in the numerator and denominator to reach the answer.