If sinθ = \(\frac{a^2 + b^2}{\sqrt {2a^4 + 2b^4}}\)
Find the value of cosec (90 - θ ) × tan θ
Answer & explanation
Correct answer: option 1
sin θ = \(\frac{P}{H}\) = \(\frac{a^2 + b^2}{\sqrt {2a^4 + 2b^4}}\)
then → B = a2 - b2
Now → cosec(90 - θ ) × tan θ
= sec θ × tan θ
= \(\frac{HP}{B^2}\)
= \(\frac{\sqrt {2a^4 + 2b^4}(a^2+b^2)}{(a^2 - b^2)^2}\)