Trigonometry Identities: Sure Questions (Class 10, Part 1)

This lesson works through a set of likely Class 10 board questions on trigonometric identities. The recurring strategy is to rewrite every ratio in terms of $\sin$ and $\cos$, take common denominators where needed, and lean on the Pythagorean identity $\sin^2 A + \cos^2 A = 1$.

Express sin A, cos A and tan A in terms of cot A

Start from $\tan A = \dfrac{1}{\cot A}$.

Sine. Using $\csc^2 A = 1 + \cot^2 A$,

$\csc A = \sqrt{1+\cot^2 A}, \qquad \sin A = \frac{1}{\sqrt{1+\cot^2 A}}.$

Cosine. From $\cos A = \sqrt{1-\sin^2 A}$,

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

And $\tan A = \dfrac{1}{\cot A}$ as stated. So all three ratios are now in terms of $\cot A$.

Prove 9 sec squared A minus 9 tan squared A equals 9

$\text{LHS} = 9\sec^2 A - 9\tan^2 A = 9\,(\sec^2 A - \tan^2 A).$

Since $\sec^2 A - \tan^2 A = 1$,

$= 9\,(1) = 9 = \text{RHS}.$

Prove the product of three brackets equals 2

Prove that

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

Write $\tan\theta = \tfrac{\sin\theta}{\cos\theta}$, $\sec\theta = \tfrac{1}{\cos\theta}$, $\cot\theta = \tfrac{\cos\theta}{\sin\theta}$ and $\csc\theta = \tfrac{1}{\sin\theta}$.

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

Take a common denominator inside each bracket:

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

Group $\sin\theta + \cos\theta$ as one block: the numerator is of the form $(a+1)(a-1) = a^2 - 1$ with $a = \sin\theta + \cos\theta$.

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

Expand $(\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + 2\sin\theta\cos\theta$, so the $1$ and the $-1$ cancel:

$= \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2 = \text{RHS}.$

Two multiple choice simplifications

Simplify (sec A + tan A)(1 - sin A)

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

The numerator is $1 - \sin^2 A = \cos^2 A$, so

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

The answer is $\cos A$.

Simplify (1 + tan squared A) over (1 + cot squared A)

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

The answer is $\tan^2 A$.

Prove cos A over (1 + sin A) plus its flip equals 2 sec A

Prove that

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

The common denominator is $(1+\sin A)\cos A$.

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

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

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

Keep the denominator factored, take $2$ common in the numerator and cancel $(1+\sin A)$:

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

Prove sin A over (1 + cos A) plus its flip equals 2 cosec A

Prove that

$\frac{\sin A}{1+\cos A} + \frac{1+\cos A}{\sin A} = 2\csc A.$

The common denominator is $(1+\cos A)\sin A$.

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

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

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

Take $2$ common and cancel $(1+\cos A)$:

$= \frac{2\,(1+\cos A)}{(1+\cos A)\sin A} = \frac{2}{\sin A} = 2\csc A = \text{RHS}.$

Key takeaways

• To put every ratio in terms of $\cot A$, build from $\csc^2 A = 1 + \cot^2 A$ and then $\sin^2 A + \cos^2 A = 1$.
• Grouping $\sin\theta + \cos\theta$ as a single block lets you apply $(a+1)(a-1) = a^2 - 1$ and cancel using $(\sin\theta+\cos\theta)^2 = 1 + 2\sin\theta\cos\theta$.
• For symmetric fraction identities, take the common denominator, keep it factored, and cancel the repeated factor after the numerator simplifies.