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If secx = 7x and tanx =\(\frac{7}{x}\) find the value of 98(x2-\(\frac{1}{x^2}\))+2. |
7 4 0 6 |
4 |
We know that sec2x - tan2x = 1 (7x)2 - (\(\frac{7}{x}\))2=1 49x2 -\(\frac{49}{x^2}\) = 1 49(x2-\(\frac{1}{x^2}\)) = 1 x2-\(\frac{1}{x^2}\)=\(\frac{1}{49}\) ⇒ Put in find ⇒ 98(\(\frac{1}{49}\))+2 = 4 |
