Trigonometry Proofs from Exercise 8.3 (Class X)

This lesson works through two trigonometric identity proofs from Class 10, Exercise 8.3. Both start from the left side and reshape it step by step until it matches the right side.

Proof 1: a square-root identity

We want to prove

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

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

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

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

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

Taking the square root of the perfect squares, $\sqrt{(1+\sin A)^2} = 1+\sin A$ and $\sqrt{\cos^2 A} = \cos A$:

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

Split the fraction into two terms:

$= \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}.$

Proof 2: a sum of two squares

We want to prove

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

Start from the left side and expand each bracket with $(a+b)^2 = a^2 + 2ab + b^2$.

Left bracket

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

Since $\sin A\,\csc A = 1$, the middle term is $2$:

$= \sin^2 A + 2 + \csc^2 A.$

Right bracket

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

Since $\cos A\,\sec A = 1$, the middle term is again $2$:

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

Add the two parts

$\text{LHS} = (\sin^2 A + \cos^2 A) + 2 + 2 + \csc^2 A + \sec^2 A.$

Use $\sin^2 A + \cos^2 A = 1$, and replace the remaining squares using $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$:

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

Collect the constants $1 + 4 + 1 + 1 = 7$:

$= 7 + \tan^2 A + \cot^2 A = \text{RHS}.$

Key takeaways

• To simplify a square root of a fraction in $\sin A$, multiply top and bottom by the conjugate so the denominator becomes $\cos^2 A$.
• The products $\sin A\,\csc A = 1$ and $\cos A\,\sec A = 1$ turn the cross terms in an expansion into simple constants.
• The identities $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$ convert leftover squares into the $\tan^2 A$ and $\cot^2 A$ wanted in the answer.