Question

plz koi sahi answer de do mithai khila dungi...XD ... but no galat answers plz ..no masti during maths plz tell... 23sub part 1 of ex 17 ...book name understanding icsc book jaha se bhi jhapna hae jhapo but tell the correct answer ​

Answer 1

Answer:

17/19

Step-by-step explanation:

5 cosθ - 12 sinθ = 0

5 cosθ = 12 sinθ

sinθ / cosθ = 5/12

tan θ = 5/12

We know,

1 + (tan^2)θ = (sec^2)θ

(sec^2)θ = 1 + (5/12)^2

==> (144+25) / 144

==> 169 / 144

sec θ = 13 / 12     .....(Taking square root)

therefore.

cos θ = 1 / sec θ => 12 / 13

(sin^2)θ + (cos^2)θ = 1 ......(Identity)

sin^2)θ + (12/13)^2 = 1

(sin^2)θ = 1 - (144/169)

==> (169-144)/169

==> 25 / 169

Thus,

(sin^2)θ => 25/169

sin θ = 5 / 13    ...(Taking square root)

sinθ + cosθ = 5/13 + 12/13 = 17/13

2.cosθ - sinθ = 2(12/13) - 5/13

==> 24/13 - 5/13

==> 19/13

Thus,

$\frac{sin.(theta)+cos.(theta)}{2.cos.(theta)-sin.(theta)} = \frac{\frac{17}{13} }{\frac{19}{13} }$

==> (17 / 13) x (13 / 19) ==> 17 / 19

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Once again thank you !!!!

ARRE MITHAI BHI KHILADO !!