Trigonometric Identity Proofs (Exercise 8.3, Part 3)

This lesson proves several identities from Class 10 Exercise 8.3. The strategy throughout is to write tangent, cotangent, secant, and cosecant in terms of $\sin\theta$ and $\cos\theta$, then simplify using standard identities.

Identity with tangent and cotangent

Prove that

$\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=1+\sec\theta\,\csc\theta.$

Write $\tan\theta=\tfrac{\sin\theta}{\cos\theta}$ and $\cot\theta=\tfrac{\cos\theta}{\sin\theta}$. The first term becomes

$\frac{\tfrac{\sin\theta}{\cos\theta}}{1-\tfrac{\cos\theta}{\sin\theta}}=\frac{\sin\theta}{\cos\theta}\cdot\frac{\sin\theta}{\sin\theta-\cos\theta}=\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)},$

and the second, after writing $\cos\theta-\sin\theta=-(\sin\theta-\cos\theta)$,

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

Taking $\tfrac{1}{\sin\theta-\cos\theta}$ common,

$\frac{1}{\sin\theta-\cos\theta}\left(\frac{\sin^2\theta}{\cos\theta}-\frac{\cos^2\theta}{\sin\theta}\right)=\frac{1}{\sin\theta-\cos\theta}\cdot\frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta}.$

Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=\sin\theta$, $b=\cos\theta$,

$\sin^3\theta-\cos^3\theta=(\sin\theta-\cos\theta)\big(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta\big).$

Cancelling $\sin\theta-\cos\theta$ and using $\sin^2\theta+\cos^2\theta=1$,

$\frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}+1=\sec\theta\,\csc\theta+1.$

So the left side equals $1+\sec\theta\,\csc\theta$.

Square root identity for secant plus tangent

Prove that

$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A.$

Multiply inside the root by $\tfrac{1+\sin A}{1+\sin A}$:

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

Proof using the difference of cubes

Prove that

$\frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}=\tan\theta.$

Take $\sin\theta$ out of the numerator and $\cos\theta$ out of the denominator:

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

Replace $\cos^2\theta=1-\sin^2\theta$ in the denominator factor:

$2\cos^2\theta-1=2(1-\sin^2\theta)-1=1-2\sin^2\theta.$

Now numerator and denominator both contain $1-2\sin^2\theta$, which cancels:

$\frac{\sin\theta}{\cos\theta}=\tan\theta.$

Expanding a sum of squares

Prove that

$(\sin A+\csc A)^2+(\cos A+\sec A)^2=7+\tan^2 A+\cot^2 A.$

Expand each square with $(a+b)^2=a^2+2ab+b^2$:

$(\sin A+\csc A)^2=\sin^2 A+2\sin A\csc A+\csc^2 A=\sin^2 A+2+\csc^2 A,$

$(\cos A+\sec A)^2=\cos^2 A+2\cos A\sec A+\sec^2 A=\cos^2 A+2+\sec^2 A.$

Adding and using $\sin^2 A+\cos^2 A=1$, $\csc^2 A=1+\cot^2 A$, and $\sec^2 A=1+\tan^2 A$:

$1+4+(1+\cot^2 A)+(1+\tan^2 A)=7+\tan^2 A+\cot^2 A.$

Chain of three equal expressions

Prove that

$\frac{1+\tan^2 A}{1+\cot^2 A}=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan^2 A.$

First equals third. Using $1+\tan^2 A=\sec^2 A$ and $1+\cot^2 A=\csc^2 A$,

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

Second equals third. Rewrite in sine and cosine:

$\left(\frac{1-\tan A}{1-\cot A}\right)^2=\frac{\left(1-\tfrac{\sin A}{\cos A}\right)^2}{\left(1-\tfrac{\cos A}{\sin A}\right)^2}=\frac{\tfrac{(\cos A-\sin A)^2}{\cos^2 A}}{\tfrac{(\sin A-\cos A)^2}{\sin^2 A}}=\frac{(\cos A-\sin A)^2}{\cos^2 A}\cdot\frac{\sin^2 A}{(\sin A-\cos A)^2}.$

Since $(\cos A-\sin A)^2=(\sin A-\cos A)^2$, those factors cancel, leaving

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

Both the first and second expressions equal $\tan^2 A$, so all three are equal.

Key takeaways

• Rewriting tangent, cotangent, secant, and cosecant as sine and cosine turns most proofs into routine algebra.
• The identities $\sin^2\theta+\cos^2\theta=1$, $\sec^2 A=1+\tan^2 A$, and $\csc^2 A=1+\cot^2 A$ do most of the simplifying work.
• Factoring tools like $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $(\cos A-\sin A)^2=(\sin A-\cos A)^2$ let common factors cancel cleanly.