PYQ - Trigonometric identitites — task. Maths CBSE Live product, Class 10 (2026-27).

If \(p\,\text{sec}\, \alpha + q\tan \alpha = x\) and \(q\,\text{sec}\, \alpha + p\tan \alpha = y\), show that \(p^2 + y^2 = q^2 + x^2\).

Proof:

x = p \sec α + q \tan α

y = p \sec α + q \tan α

Squaring both sides of the equations, we get

---------\((1)\)

----------\((2)\)

Subtracting equation \((2)\) from \((1)\), we get

This implies that,

Rearranging the terms,

Hence proved.

Answer variants:

p^{2} (1) - q^{2} (1) = x^{2} - y^{2}

p^{2} + y^{2} = q^{2} + x^{2}

p^{2} \sec^{2} α + q^{2} \tan^{2} α + 2 pq \sec α \tan α = x^{2}

q^{2} \sec^{2} α + p^{2} \tan^{2} α + 2 pq \sec α \tan α = y^{2}

\sec^{2} α + \tan^{2} α = 1

p^{2} \sec^{2} α + q^{2} \tan^{2} α - 2 pq \sec α \tan α - q^{2} \sec^{2} α - p^{2} \tan^{2} α + 2 pq \sec α \tan α = x^{2} - y^{2}

a^{2} \sec^{2} θ - b^{2} \tan^{2} θ + 2 ab \sec θ \tan θ - b^{2} \sec^{2} θ - a^{2} \tan^{2} θ - 2 ab \sec θ \tan θ = m^{2} - n^{2}

\sec^{2} α - \tan^{2} α = 1

Did you find an error?

7. PYQ - Trigonometric identitites