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Q) Prove that: (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2.

(Q 29 B – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Let’s start from LHS:

∴ LHS = (sin A + sec A) 2 + (cos A + cosec A) 2

By algebraic identity, we have: (a + b) 2 = a 2 + b 2 + 2 a b

∴ LHS = (sin A + sec A) 2 + (cos A + cosec A) 2

∴ LHS = (sin 2 A + sec 2 A + 2 sin A sec A) + (cos 2 A + cosec 2 A + 2 cos A cosec A)

∴ LHS = (sin 2 A + \frac{1}{\cos^2 A} + 2 sin A \frac{1}{\cos A}) + (cos 2 A + \frac{1}{\sin^2 A} + 2 cos A \frac{1}{\sin A})

∴ LHS = sin 2 A + \frac{1}{\cos^2 A} + 2 \frac{\sin A }{\cos A} + cos 2 A + \frac{1}{\sin^2 A} + 2 \frac{\cos A }{\sin A}

∴ LHS = sin 2 A + cos 2 A + \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} + 2 \frac{\sin A }{\cos A} + 2 \frac{\cos A}{\sin A}

∴ LHS = sin 2 A + cos 2 A + \frac{\sin^2 A + \cos^2 A}{\sin^2 A . \cos^2 A} + 2 \frac{\sin^2 A + \cos^2 A}{\sin A . \cos A}

By trigonometric identity, we have sin 2 θ + cos 2 θ = 1

∴ LHS = sin 2 A + cos 2 A + \frac{\sin^2 A + \cos^2 A}{\sin^2 A . \cos^2 A} + 2 \frac{\sin^2 A + \cos^2 A}{\sin A . \cos A}

∴ LHS = 1 + \frac{1}{\sin^2 A . \cos^2 A} + 2 \frac{1}{\sin A . \cos A}

∴ LHS = 1 + \frac{1}{\sin^2 A . \cos^2 A} + \frac{2}{\sin A . \cos A}

∴ LHS = 1 + \frac{1 + 2 \sin A . \cos A}{\sin^2 A . \cos^2 A}

∴ LHS = \frac{\sin^2 A . \cos^2 A + 1 + 2 \sin A . \cos A}{\sin^2 A . \cos^2 A}

If we consider, sin A . cos A = P, then:

∴ LHS = \frac{P^2+ 1 + 2 P}{\sin^2 A . \cos^2 A}

∴ LHS = \frac{(P + 1)^2}{\sin^2 A . \cos^2 A}         (∵ (a + b) 2 = a 2 + b 2 + 2 a b)

∴ LHS = (\frac{(P + 1)}{(\sin A . \cos A)})^2

Let’s substitute back the value of P = sin A . cos A in LHS:

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

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

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

∴ LHS = (1 + cosec A . sec A) 2

∴ LHS = (1 + sec A . cosec A) 2 = RHS

Hence Proved !

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