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If sinθ = \(\frac{a^2 + b^2}{\sqrt {2a^4 + 2b^4}}\) Find the value of cosec (90 - θ ) × tan θ |
\(\frac{\sqrt {2a^4 + 2b^4}(a^2+b^2)}{(a^2 - b^2)^2}\) \(\frac{\sqrt {a^2 - b^2} + (a+b)}{2\sqrt {ab}}\) 0 \(\frac{1}{\sqrt {ab}}\) × \(\frac{\sqrt {a^2 - b^2}}{4ab^2}\) |
\(\frac{\sqrt {2a^4 + 2b^4}(a^2+b^2)}{(a^2 - b^2)^2}\) |
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 = \(\frac{HP}{B^2}\) = \(\frac{\sqrt {2a^4 + 2b^4}(a^2+b^2)}{(a^2 - b^2)^2}\) |
