Math, asked by pubgthrillers1209, 1 year ago


Trigonometry question in photo 10 grade cbse question

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Answers

Answered by UltimateMasTerMind
10

Solution:-

Given:-

tanθ = 20/21

To Prove:-

[ ( 1-sinθ + cosθ) / ( 1 + sinθ + cosθ ) ] = 3/7

Proof :-

Let the sides of the Triangle be "x".

We know that,

tanθ = Perpendicular/ Base

=) tanθ = Perpendicular/Base = 20/21

Hence,

Perpendicular = 20x

Base = 21x

By Pythagoras Theorem,

Hypotenuse² = Perpendicular² + Base²

=) H² = ( 20x )² + ( 21x )²

=) H² = 400x² + 441x²

=) H = √841x²

=) H = 29x

Hence,

Hypotenuse = 29x

Now, Finding the value of sinθ and cosθ.

sinθ = Perpendicular/Hypotenuse

=) sinθ = 20x / 29x

=) sinθ = 20/29

cosθ = Base / Hypotenuse

=) cosθ = 21x / 29x

=) cosθ = 21/29

Substituting the values of "sinθ" and "cosθ" in the Equation. We get,

=) [ ( 1 - 20/29 + 21/29 ) / ( 1 + 20/29 + 21/29 ) ]

=) [ ( 29 - 20 + 21 ) / ( 29 + 20 + 21 ) ]

=) 30/70

=) 3/7

Hence Proved!

Answered by tejasgupta
10

Answer:

No answer. Just check the explanation...

Step-by-step explanation:

Let θ = A.

Then, tanA = Perp./Base = 20/21

Let P = 20k and B = 21k.

Now, By Pythagoras theorem,

H = √k²(20²+ 21²) = k√400+441 = 29k

sinA = P/H = 20k/29k = 20/29

cosA = B/H = 21k/29k = 21/29

To Prove:-

(1 - sinA + cosA)/(1 + sinA + cosA) = 3/7

Taking LHS:

(1 - sinA + cosA)/(1 + sinA + cosA)

= (1 - 20/29 + 21/29)/(1 + 20/29 + 21/29)

= (29 - 20 + 21/29)(29 + 20 + 21/29)

= (50-20)/(70)

= 30/70

= 3/7

= RHS

LHS = RHS

Hence Proved.

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