Question:
Let \(a\sin^2 x + b\cos^2 x = c\), \(b\sin^2 y + a\cos^2 y = d\) and p tan x = q tan y. What is \(\dfrac{d-a}{b-d}\) equal to?
मान लीजिए \(a\sin^2 x + b\cos^2 x = c\), \(b\sin^2 y + a\cos^2 y = d\) और \(p\tan x = q\tan y\) है। \(\dfrac{d-a}{b-d}\) किसके बराबर है?
Options:
- \(\sin^2 y\)
- \(\tan^2 y\)
- \(\cos^2 y\)
- \(\cot^2 y\)
- \(\sin^2 y\)
- \(\tan^2 y\)
- \(\cos^2 y\)
- \(\cot^2 y\)
Correct Answer: b)
Explanation:
Dividing \(b\sin^2 y + a\cos^2 y = d\) by \(\cos^2 y\):
\(b\tan^2 y + a = d(1+\tan^2 y) \implies (b-d)\tan^2 y = d-a \implies \dfrac{d-a}{b-d} = \tan^2 y\).
\(b\sin^2 y + a\cos^2 y = d\) को \(\cos^2 y\) से भाग देने पर:
\(b\tan^2 y + a = d(1+\tan^2 y) \implies \dfrac{d-a}{b-d} = \tan^2 y\)।