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Kumar D.
Tutor for six years
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Trigonometry
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Question:

Question : Prove the identity $$\frac{1+cos(x)+cos(2x)}{sin(x)+sin(2x)}$$ = $$cot(x)$$

Kumar D.

Solution: We know $$cos(2x)= 2cos^2(x)-1$$ and $$sin(2x)=2sin(x)cos(x)$$ equation $$\frac{1+cos(x)+cos(2x)}{sin(x)+sin(2x)}$$ can be rewrite as $$\frac{1+cos(x)+2cos^2(x)-1}{sin(x)+2sin(x)cos(x)}$$ = $$\frac{cos(x)+2cos^2(x)}{sin(x)+2sin(x)cos(x)}$$ = $$\frac{cos(x)[1+2cos(x)]}{sin(x)[1+2cos(x)]}$$ = $$cot(x)$$ $$\because cot(x)$$ = $$\frac{cos(x)}{sin(x)}$$

Basic Math
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Question:

Question : Solve using quadratic formula. $$2g^2$$ - 3$$g$$ - 2 = 0

Kumar D.

Solution : Quadratic formula standard equation : $$ax^2 + bx + c$$=0 : Roots = $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ here $$a=2$$, $$b= -3$$, and $$c= -2$$ putting these values in the formula Roots = $$\frac{-(-3)\pm\sqrt{(-3)^2-4*2*(-2)}}{2*2}$$ {$$\because {(-)*(-) = (+)}$$} or, $$\frac{3\pm\sqrt{9-(-16)}}{4}$$ or $$\frac{3\pm\sqrt{9+16}}{4}$$ {$$\because {(-)*(-) = (+)}$$} or, $$\frac{3\pm\sqrt{25}}{4}$$ or, $$\frac{3\pm5}{4}$$ {$$\sqrt{25}=5$$} Roots are $$2$$ (if we select positive sign) and $$\frac{-1}{2}$$ (negative sign is selected) or, $$2$$ and $$\frac{-1}{2}$$

Algebra
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Question:

Question 1 : If $$x+y$$ is constant, prove that $$xy$$ is maximum when $$x=y$$.

Kumar D.

Answer: We can write $$xy$$ as $$\frac{(x+y)^2}{2}$$ - $$\frac{(x-y)^2}{2}$$ We know that the square of any real no. $$\geq$$ 0 So, $$\frac{(x+y)^2}{2}$$ $$\geq$$ 0 and $$\frac{(x-y)^2}{2}$$ $$\geq$$ 0 As $$x+y$$ is constant, xy is the maximum when $$\frac{(x-y)^2}{2}$$ = 0 or, $$\frac{x-y}{2}$$ = 0 or, $$x-y$$ = 0 or, $$x=y$$ hence proved.

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