Question

do it fast .plz
give only right answer.otherwise i will report it.​

Answer 1

Solution :-

Concepts used to solve problem :-

• Trigonometric ratios Complementary angles
• Trigonometric identities

$\dfrac{ \sin^{2}63 + \sin^{2}27 }{ \cos^{2} 17 + \cos^{2} 73}$

It can be written as

$= \dfrac{ \sin^{2}63 + \sin^{2}(90 - 63) }{ \cos^{2} (90 - 73) + \cos^{2} 73}$

$= \dfrac{ \sin^{2}63 + \cos^{2}63 }{ \sin^{2} 73 + \cos^{2} 73}$

[ Because sin(90 - θ) = cosθ and cos(90 - θ) = sin θ ]

$= \dfrac{1 }{1} = 1$

[ Because sin² θ + cos² θ = 1 ]

Therefore the value is 1.

Answer 2

Answer:

1

Step-by-step explanation:

Sin²63° + Sin²27° ÷ Cos²17° + Cos²73°

BY CHANGING ANY ONE TRIGONOMETRIC RATIO IN BOTH NUMERATOR AND DENOMINATOR WE GET

Sin²63° + Sin² (90° - 63°) ÷

Cos² (90° - 73°) + Cos²73°

= Sin²63° + Cos²63° ÷ Sin²73° + Cos²73°

REASON :- [ Sin ( 90° - Φ ) = CosΦ &

vice versa ]

= 1/1 REASON :- [(Sin²Φ + Cos²Φ) = 1]

= 1