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Question 140
If $$\frac{2tan^230^{\circ}}{1-tan^230^{\circ}}+sec^{2}45^{\circ}-sec^{2}0^{\circ}$$ = x sec 60°, then the value of x is
Solution
we know ,
tan 30 = $$\frac{1}{\surd3}$$
sec 45 = $$\surd2$$
sec 0 = 1
sec 60 = 2
using the above values in
L.H.S :: $$\frac{2tan^230^{\circ}}{1-tan^230^{\circ}}+sec^{2}45^{\circ}-sec^{2}0^{\circ}$$ = $$\frac{2 (\frac{1}{\surd3})^2}{1 - (\frac{1}{\surd3})^2}$$ + $$(\surd2)^2$$ - $$1^2$$
= 1 + 2 - 1 = 2
it is given that ,
2 = x sec 60 = 2x
x = 1
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