If Tan A Tan B + \(\frac{{\cos x}}{{\cos A\cos B}}\) = 1, then x = ?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 08 Dec 2022 Shift 4)

Answer 1

Given :

Tan A Tan B + \(\frac{{\cos x}}{{\cos A\cos B}}\) = 1

Formula used:

\(Tan\:\theta={Sin\:\theta\over Cos\:\theta}\)

Cos (a + b)= cos a cos b - sin a sin b

Calculation:

Tan A Tan B + \(\frac{{\cos x}}{{\cos A\cos B}}\) = 1

⇒ Tan A Tan B Cos A Cos B + Cos x = Cos A Cos B

⇒ Sin A Sin B + Cos x = Cos A Cos B

⇒ Cos x = Cos A Cos B - Sin A Sin B

⇒ Cos x = Cos (A + B)

⇒ x = A + B

Hence, 'Option 3' is the correct answer.

Shortcut Trick

If we put A = 45° and B = 45°, we get

Tan A Tan B + \(\frac{{\cos x}}{{\cos A\cos B}}\) = 1

⇒ 1 + \(\frac{cos x}{(\frac{1}{\sqrt2})^2}\) = 1

⇒ 1 + 2cos x = 1

⇒ cos x = 0

so, x = 90° = A + B