Sin θ = 20/29 then Sin2 θ + cos2 θ =A) 144/841B) 41/841C) 1D) 0

1.	Sin θ = 20/29 then Sin2 θ + cos2 θ =A) 144/841B) 41/841C) 1D) 0
Answer» Correct option is: C) 1 \(Sin^2 \theta + cos^2 \theta\) = 1 for any value of \(\theta\). So, hence if sin \(\theta\) = \(\frac {20}{29}\) then \(Sin^2 \theta + cos^2 \theta\) = 1 Alternative : We have sin \(\theta\) =\(\frac {20}{29}\) \(\therefore\) \(cos^2\, \theta = 1- sin^2 \theta = 1 - (\frac {20}{29})^2 = 1- \frac {400}{841}\) = \(\frac {841-400}{841} = \frac {441}{841}\) Now, \(Sin^2 \theta + cos^2 \theta\) = \((\frac {20}{29})^2+ \frac {441}{841} = \frac {400}{841} + \frac {441}{841} = \frac {841}{841} = 1\) Correct option is: C) 1

Answer»

Correct option is: C) 1

\(Sin^2 \theta + cos^2 \theta\) = 1 for any value of \(\theta\).

So, hence if sin \(\theta\) = \(\frac {20}{29}\) then \(Sin^2 \theta + cos^2 \theta\) = 1

Alternative :

We have sin \(\theta\) =\(\frac {20}{29}\)

\(\therefore\) \(cos^2\, \theta = 1- sin^2 \theta = 1 - (\frac {20}{29})^2 = 1- \frac {400}{841}\)

= \(\frac {841-400}{841} = \frac {441}{841}\)

Now, \(Sin^2 \theta + cos^2 \theta\) = \((\frac {20}{29})^2+ \frac {441}{841} = \frac {400}{841} + \frac {441}{841} = \frac {841}{841} = 1\)

Correct option is: C) 1

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