In ΔABC, right angled at B, if sin A = \(\frac{1}{\sqrt2}\), then t

1.	In ΔABC, right angled at B, if sin A = \(\frac{1}{\sqrt2}\), then the value of \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) is:1. \(2 \sqrt{5}\)2. 13. 34. 2
Answer» Correct Answer - Option 2 : 1 Given :- ΔABC is a right angle at B sin A = (1/√2) Concept :- As sin A = (1/√2) sin A = sin45° A = 45° Calculation :- As B is right angle and, ⇒ ∠A = 45° Sum of triangle = 180° ⇒ ∠A + ∠B + ∠C = 180° ⇒ 45° + 90° + ∠C = 180° ⇒ ∠C = 180° - 135° ⇒ ∠C = 45° As ∠A = ∠C = 45° ⇒ sin A = cos C = cos A = (1/√2) ΔABC is an isosceles triangle Now, Put the value of all identities ⇒ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = (sin A (sin A + sin A))/(sin A (sin A + sin A)) ⇒ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = 1 ∴ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = 1

In ΔABC, right angled at B, if sin A = \(\frac{1}{\sqrt2}\), then the value of \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) is:1. \(2 \sqrt{5}\)2. 13. 34. 2

Answer» Correct Answer - Option 2 : 1

Given :-

ΔABC is a right angle at B

sin A = (1/√2)

Concept :-

As sin A = (1/√2)

sin A = sin45°

A = 45°

Calculation :-

As B is right angle and,

⇒ ∠A = 45°

Sum of triangle = 180°

⇒ ∠A + ∠B + ∠C = 180°

⇒ 45° + 90° + ∠C = 180°

⇒ ∠C = 180° - 135°

⇒ ∠C = 45°

As ∠A = ∠C = 45°

⇒ sin A = cos C = cos A = (1/√2)

ΔABC is an isosceles triangle

Now, Put the value of all identities

⇒ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = (sin A (sin A + sin A))/(sin A (sin A + sin A))

⇒ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = 1

∴ \(\frac{sin A(cos C + cos A)}{cos C (sin C + sin A)}\) = 1