Question

Here's a Math question from the Chapter Trigonometry.
The question is provided in the attachment.
Solve with proper explanation.
Expecting to see good answers !
Don't think of copying...

Answer 1

Answer:

Step-by-step explanation:

Refer to attachment for Diagram.

According to Diagram:

Sin²θ = ( AB / AC )²

Sin⁴θ = ( AB / AC )⁴

Cos²θ = ( BC / AC )²

Cos⁴θ = ( BC / AC )⁴

Similarly, for Ф we can write it as:

Sin²Ф = ( BC / AC )²

Sin⁴Ф = ( BC / AC )⁴

Cos²Ф = ( AB / AC )²

Cos⁴Ф = ( AB / AC )⁴

Now According to the question,

( Sin²θ / Cos⁴Ф ) + ( Sin⁴Ф / Cos²θ  ) = 1

⇒ [ ( AB / AC )² / ( AB / AC )⁴ ] + [ ( BC / AC )⁴ / ( BC / AC )² ] = 1

⇒ [ AB² / AC² / AB⁴ / AC⁴ ] + [ BC⁴ / AC⁴ / BC² / AC² ] = 1

⇒ AC² / AB² + BC² / AC² = 1   ... ( Eqn. 1 )

Similarly,

Cos⁴θ / Sin²Ф + Cos²Ф / Sin⁴θ

Substituting the values we get,

⇒ [ ( BC / AC )⁴ / ( BC / AC )² ] + [ ( AB / AC )² / ( AB / AC )⁴

⇒ [ BC⁴ / AC⁴ / BC² / AC² ] + [ AB² / AC² / AB⁴ / AC⁴ ]

⇒ BC² / AC² + AC² / AB²

This is the same as we got in Equation 1.

Hence this is equal to 1.

Hence, Cos⁴θ / Sin²Ф + Cos²Ф / Sin⁴θ = 1

Hence Proved !!

Hope it helped !!