# Prove That: ( Sin 49 ∘ Cos 41 ∘ ) 2 + ( Cos 41 ∘ Sin 49 ∘ ) 2 = 2 - Mathematics

Sum

Prove that:

$\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2$

#### Solution

$LHS = \left( \frac{\sin49°}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\sin49°} \right)^2$

$= \left( \frac{\cos\left( 90° - 49° \right)}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\cos\left( 90° - 49° \right)} \right)^2$

$= \left( \frac{\cos41°}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\cos41°} \right)^2$

= 12 + 12

= 1 + 1

= 2

= RHS

Concept: Trigonometric Ratios of Complementary Angles
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#### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 6.5 | Page 313
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