Math, asked by rajraut456456, 8 months ago

evaluate (sin27° ÷ cos 63° )2 - (cos63° ÷ sin 27°)2​

Answers

Answered by DibyenduChakraborty2
3

(sin27° ÷ cos 63° )² - (cos63° ÷ sin 27°)²

= [sin 27° / sin(90°-63°)]² - [cos 63° / cos(90°-27°)]²

= [sin 27°/sin 27°]² - [cos 63°/cos 63°]²

= 1² - 1²

= 1 - 1

= 0.

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