Question

cos square theta by 1 minus tan theta + sin cube theta by sin theta minus cos theta is equal to 1 + sin theta cos theta

Answer 1

Hey there !!
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Here we have


From LHS -

{ Cos ²∅ / 1 - sin∅/ cos∅ } + { sin ³∅ / sin∅ - cos ∅ }


⇒ { Cos² ∅ × cos∅/ cos∅- sin ∅ } + { sin ³∅/ sin ∅ - cos∅ }

⇒. { cos³∅ / - ( sin ∅ - cos∅ ) } + { sin³∅ / sin ∅ - cos∅ }

⇒. cos³∅ - sin³∅ / sin ∅ - cos∅


{ Using Formula - a³ - b³ ( a - b ) ( a² + b² + ab }

Now ,

⇒ { ( sin∅ - cos∅ ) ( sin ²∅ + cos² ∅+ sin∅. × cos∅ )} / sin∅ - cos ∅

We know sin²∅ + cos²∅ = 1

Then ,

➯ 1 + sin ∅. Cos∅

Hence proved



Thanks!

Answer 2

Answer:

Step-by-step explanation:

Cos2theta/ (1- tan theta) + Sin3theta/(sin theta - cos theta)

Cos2theta [1 - (sin theta/cos theta)] - Sin3theta/(Cos theta - Sin theta)

Cos2theta . Cos theta/(Cos theta - Sin theta) - Sin3theta/(Cos theta - Sin theta)

Cos3theta /(Cos theta - Sin theta) - Sin3theta/(Cos theta - Sin theta)

(Cos3theta - Sin3theta)/(Cos theta - Sin theta)

{a3 - b3} = (a - b) (a2 + ab + b2)

on applying above formula in numerator, we get:

(Cos theta - Sin theta)(Cos2theta + Sin2theta + Cos theta.Sin theta) /(Cos theta - Sin theta)

(Cos2theta + Sin2theta + Cos theta.Sin theta)

(1 + Cos theta.Sin theta) = R.H.S