Math, asked by L12345, 1 year ago

please solve this question friends

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Answered by siddhartharao77
3
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!

siddhartharao77: is it right or wrong?
siddhartharao77: Thanks Sis..
Answered by Anonymous
6
Hi,

Please see the attached file!


Thanks
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L12345: WHAT PERCENT DID YOU GOT IN CLASS 10TH BOARD RESULTS
siddhartharao77: Nice explanation...
Anonymous: Thank you Bhai ! 10 CGPA @L12345
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