Trigonometric Identity Proofs (Exercise 8.3, Part 2)

This lesson works through four proofs from Exercise 8.3 of the Class 10 trigonometry chapter. An identity can be proved by simplifying the left side down to the right, the right side down to the left, or by reducing both sides to a common expression.

Proving $\sqrt{\tfrac{1+\sin a}{1-\sin a}} = \sec a + \tan a$

Start from the left side and rationalise by multiplying numerator and denominator inside the root by the conjugate of the denominator, $1 + \sin a$.

$\sqrt{\frac{1+\sin a}{1-\sin a}} = \sqrt{\frac{(1+\sin a)(1+\sin a)}{(1-\sin a)(1+\sin a)}} = \sqrt{\frac{(1+\sin a)^2}{1-\sin^2 a}}$

Since $1 - \sin^2 a = \cos^2 a$, the root simplifies cleanly.

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

This equals the right side.

Proving $(\sin a + \csc a)^2 + (\cos a + \sec a)^2 = 7 + \tan^2 a + \cot^2 a$

Take the left side and expand each bracket with $(x+y)^2 = x^2 + 2xy + y^2$.

$\sin^2 a + \csc^2 a + 2\sin a\,\csc a + \cos^2 a + \sec^2 a + 2\cos a\,\sec a$

Here $\sin a\,\csc a = 1$ and $\cos a\,\sec a = 1$, and $\sin^2 a + \cos^2 a = 1$.

$1 + \csc^2 a + \sec^2 a + 2 + 2$

Now replace $\csc^2 a = 1 + \cot^2 a$ and $\sec^2 a = 1 + \tan^2 a$ to bring in the terms needed on the right.

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

This equals the right side.

Proving $\dfrac{1+\tan^2 a}{1+\cot^2 a} = \left(\dfrac{1-\tan a}{1-\cot a}\right)^2 = \tan^2 a$

With three expressions, prove the first equals the last, then the middle equals the last; if both equal $\tan^2 a$, all three are equal.

First expression equals the last

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

Middle expression equals the last

Replace $\cot a = \tfrac{1}{\tan a}$ and take the LCM in the denominator.

$\left(\frac{1-\tan a}{1-\cot a}\right)^2 = \frac{(1-\tan a)^2}{\left(\frac{\tan a - 1}{\tan a}\right)^2} = \frac{(1-\tan a)^2\,\tan^2 a}{(\tan a - 1)^2}$

Since $(1-\tan a)^2 = (\tan a - 1)^2$, those factors cancel.

$= \tan^2 a$

Both pieces equal $\tan^2 a$, so all three expressions are equal.

Proving $\dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\,\csc\theta$

Write $\tan\theta$ and $\cot\theta$ in terms of sine and cosine, then simplify each denominator with an LCM.

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

Take a negative sign out of the second denominator so both brackets read $\sin\theta - \cos\theta$, then factor it out.

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

The numerator is a difference of cubes, $a^3 - b^3 = (a-b)(a^2+ab+b^2)$.

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

Cancel $\sin\theta - \cos\theta$, and use $\sin^2\theta+\cos^2\theta = 1$.

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

This equals the right side.

A second approach to a related result

The expression $\dfrac{1}{\tan a + \cot a}$ can also be simplified directly.

$\frac{1}{\tan a + \cot a} = \frac{1}{\frac{\sin a}{\cos a} + \frac{\cos a}{\sin a}} = \frac{\sin a\cos a}{\sin^2 a + \cos^2 a} = \frac{\sin a\cos a}{1} = \sin a\cos a$

Key takeaways

• Rationalising a fraction under a root with the conjugate turns $1 - \sin^2 a$ into $\cos^2 a$.
• After expanding, use $\sin a\,\csc a = 1$, $\cos a\,\sec a = 1$, $\csc^2 a = 1 + \cot^2 a$, and $\sec^2 a = 1 + \tan^2 a$.
• A three-part identity is proved by showing each expression reduces to the same value.
• Factoring out a common denominator and applying the difference of cubes simplifies the combined fraction to $1 + \sec\theta\,\csc\theta$.