please solve this question friends
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Note: Here I am writing theta as A.
Now,
Given tanA + secA = 5/3 -------- (1)
We know that tan^2A + 1 = sec^2A
tan^2A - sec^2A = -1
We know that a^2 - b^2 = (a + b)(a - b)
(tanA + secA)(tanA - secA) = -1
(5/3)(tanA - secA) = -1
tanA - secA = -3/5 --------- (2)
On solving (1) & (2), we get
tanA + secA = 5/3
tanA - secA = -3/5
---------------------------
2tanA = 16/15
tanA = 16/30. -------- (3)
Substitute (3) in (1), we get
= > (16/30) + secA = 5/3
= > secA = 5/3 - 16/30
= > secA = 5/3 - 8/15
= > secA = 17/15
we know that cosA = 1/secA
= 1/17/15
= 15/17.
Now,
We know that sin^2A + cos^2A = 1
= > sin^2A = 1 - cos^2A
= > sin^2A = 1 - (15/17)^2
= > sin^2A = 1 - 225/289
= > sin^2A = (64/289)
= > sinA = 8/17.
Therefore the value of sintheta = 8/17.
Hope this helps!
Now,
Given tanA + secA = 5/3 -------- (1)
We know that tan^2A + 1 = sec^2A
tan^2A - sec^2A = -1
We know that a^2 - b^2 = (a + b)(a - b)
(tanA + secA)(tanA - secA) = -1
(5/3)(tanA - secA) = -1
tanA - secA = -3/5 --------- (2)
On solving (1) & (2), we get
tanA + secA = 5/3
tanA - secA = -3/5
---------------------------
2tanA = 16/15
tanA = 16/30. -------- (3)
Substitute (3) in (1), we get
= > (16/30) + secA = 5/3
= > secA = 5/3 - 16/30
= > secA = 5/3 - 8/15
= > secA = 17/15
we know that cosA = 1/secA
= 1/17/15
= 15/17.
Now,
We know that sin^2A + cos^2A = 1
= > sin^2A = 1 - cos^2A
= > sin^2A = 1 - (15/17)^2
= > sin^2A = 1 - 225/289
= > sin^2A = (64/289)
= > sinA = 8/17.
Therefore the value of sintheta = 8/17.
Hope this helps!
siddhartharao77:
is it right or wrong?
Answered by
6
Hi,
Please see the attached file!
Thanks
Please see the attached file!
Thanks
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