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Arvind1234
@Arvind1234
July 2021
1
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Solve this question immediate answers required
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parisakura98pari
Given cosx = t 0 < t < 1
⇒ cosx is positive
now cosx = t = Base/ hypotense
consider a triangle with base = t and hypo. = 1
therefore perpendicular = √1-t²
now sinx = perp. / hypo = √1-t²/1
and tanx = √1-t² / t
but A/Q x ∉ 1st quadrant and cosx too positive therefore x∈ 4th quadrant
and sinx = -√1-t²/1 and tanx = -√1-t²/t
hope u'll get it if doubt then ask........
2 votes
Thanks 1
Arvind1234
Trying to catch it let me
Arvind1234
Hello
Arvind1234
I think that the first is eliminated
Arvind1234
Third and fourth quardants are eliminated due to negative values
Arvind1234
Only second is reamined
Arvind1234
Now what
parisakura98pari
yep! I just want to remind u the condition given in question.
Arvind1234
Ok
parisakura98pari
any more queries?
Arvind1234
done
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Answers & Comments
⇒ cosx is positive
now cosx = t = Base/ hypotense
consider a triangle with base = t and hypo. = 1
therefore perpendicular = √1-t²
now sinx = perp. / hypo = √1-t²/1
and tanx = √1-t² / t
but A/Q x ∉ 1st quadrant and cosx too positive therefore x∈ 4th quadrant
and sinx = -√1-t²/1 and tanx = -√1-t²/t
hope u'll get it if doubt then ask........