Sec^2 10 - cot ^280 + sin15 cos 75 + cos15 sin75 / cos theta.sin(90-theta) + sin theta.cos(90-theta) = 2
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Answer:
Step-by-step explanation:
Hi there!
We have,
sec²(10) − cot²(80) + sin(15) cos(75) + cos(15) sin(75)/ cosθ sin(90-θ) + sinθ cos(90-θ)
Let's divide these into three parts :
♦ sec²(10) − cot²(80)
♦ sin(15) cos(75) + cos(15) sin(75)
♦ cosθ sin(90-θ) + sinθ cos(90-θ)
Simplifying Each :
♦ sec²(10) − cot²(80)
= 1/cos²(10) − cos²(80)/sin²(80)
= 1/sin²(80) − cos²(80)/sin²(80)
= (1−cos²(80))/sin²(80)
= sin²(80)/sin²(80)
= 1
♦ sin(15) cos(75) + cos(15) sin(75)
= sin(15+75)
= sin(90)
= 1
♦ cosθ sin(90-θ) + sinθ cos(90-θ)
= sin [ θ+(90−θ) ]
= sin(90)
= 1
Putting all them back together :
= 1 + 1 / 1
= 1 + 1
= 2
[ Thank you! for asking the question. ]
Hope it helps!
[ NOTE : It's was a bit long question... O_O, that's why I've done it in parts ]
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