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# Tutor profile: Nishant G.

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Nishant G.
Brown University Student, Teacher for over 4 years
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## Questions

### Subject:SAT

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Question:

The following questions are concerned with the SAT writing section. Correct the following phrases: (a) After arriving late to class, a convincing excuse was needed by the student. (b) The student was asked to write the lab report diligently, precisely, and in a thorough manner.

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Nishant G.

(a) This question involves something known as a dangling modifier, which is essentially a phrase or a word that modifies or attributes an action to a word not clearly stated in the sentence. It is essential that the modifier only applies to the object it is modifying. In this sentence, the phrase that describes the action is "After arriving late to class," but the doer is not the "convincing excuse." The "student" is the doer. To correct this dangling modifier so that it applies to the object is modifying, we correct the sentence to: After arriving late to class, the student needed a convincing excuse. Now, it is clear that the student is indeed the one who arrived late to class. (b) This question deals with parallel structure in sentences, which means that the same pattern of words or phrases must be used when there are two or more ideas expressed in a sentence. In this sentence, the words "diligently" and "precisely" are in the adverb form, whereas "in a thorough manner" is a phrase. All three ideas need to be expressed in the same form, so the corrected sentence would be: The student was asked to write the lab report diligently, precisely, and thoroughly. Now, all the three words are in the adverb form, making the sentence in accordance with the parallelism rules.

### Subject:Trigonometry

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Question:

Prove: $$\frac{1 +\sin{x} + \cos{x}}{1 + \sin{x} - \cos{x}} = \cot{\frac{1}{2}x}$$.

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Nishant G.

Proving trigonometric identities is one of the most commonly asked questions in trigonometry. The problem that I have devised is not a straightforward identity to prove, rather it uses an array of other identities that students should know, such as the double angle identifies, in addition to some cheeky algebraic manipulation in order to solve. To prove this, we start from the left hand side. When ever we see a bunch of terms in the denominator of an identity, it makes sense to try and simplify it. In this case, one way to simplify $$1 + \sin{x} - \cos{x}$$ is to multiply it with $$1 + \sin{x} + \cos{x}$$ using the fact that $$(a+b)(a-b) = a^2 - b^2$$ where $$a = 1 + \sin{x}$$ and $$b = \cos{x}$$. Thus we multiply the denominator and numerator of the left hand side of the expression with $$1 + \sin{x} + \cos{x}$$. (Multiplying an expression in the numerator and denominator does not change the value of the expression as the two terms cancel out.) $$\frac{1 +\sin{x} + \cos{x}}{1 + \sin{x} - \cos{x}}$$ $$= \frac{(1 +\sin{x} + \cos{x})^2}{(1 + \sin{x} - \cos{x})(1 + \sin{x} + \cos{x})}$$ $$= \frac{(1 +\sin{x})^2 +2(1 + \sin{x}) \cos{x} + \cos^{2}x}{(1 + \sin{x})^2 - \cos^{2}x}$$ $$= \frac{1 + \sin^{2}x + 2\sin{x} + 2\cos{x} +2\sin{x}\cos{x}+ \cos^{2}x}{1 + 2\sin{x}+ \sin^2{x} - \cos^{2}x}$$ We use the fact that $$\sin^2{x} + \cos^{2}{x} = 1$$ to proceed: $$= \frac{1 + 2\sin{x} + 2\cos{x} +2\sin{x}\cos{x}+ 1}{\sin^2{x} + \cos^{2}{x} + 2\sin{x}+ \sin^2{x} - \cos^{2}x}$$ (We simplified the $$\sin^2{x} + \cos^{2}{x}$$ in the numerator and broke down the $$1$$ into $$\sin^2{x} + \cos^{2}{x}$$ in the denominator) $$= \frac{2 + 2\sin{x} + 2\cos{x} +2\sin{x}\cos{x}}{2\sin^2{x} + 2\sin{x}}$$ $$= \frac{1 + \sin{x} + \cos{x} +\sin{x}\cos{x}}{\sin^2{x} + \sin{x}}$$ $$= \frac{\sin{x}(1 + \cos{x}) +(1 + \cos{x})}{ \sin{x}(1 + \sin{x})}$$ (We factorized the numerator and denominator by taking common terms) $$= \frac{(\sin{x} + 1)(1 + \cos{x})}{ \sin{x}(1 + \sin{x})}$$ $$= \frac{(1 + \cos{x})}{ \sin{x}}$$. It seems we have succeeded in simplifying the initially complicated expression into a much simpler one. However, to proceed we will use the double angle formulas. Firstly, $$\cos{2a} = 2\cos^{2}{a} - 1 \implies cos^{2}{a} = \frac{1}{2}(\cos{2a} + 1)$$. Substituting $$a = \frac{x}{2}$$, we get $$cos^{2}{\frac{x}{2}} = \frac{1}{2}(\cos{x} + 1)$$. Let this be equation $$(i)$$ Similarly, $$\sin{2a} = 2\sin{a}\cos{a} \implies \sin{a}\cos{a} = \frac{1}{2}\sin{2a}$$ Substituting $$a = \frac{x}{2}$$, we get $$\sin{\frac{x}{2}}\cos{\frac{x}{2}} = \frac{1}{2}\sin{x}$$. Let this be equation $$(ii)$$ Multiplying the numerator and denominator of our simplified expression by $$\frac{1}{2}$$: $$= \frac{\frac{1}{2}(1 + \cos{x})}{\frac{1}{2} \sin{x}}$$. We now insert $$(i) and$$ $$(ii)$$ in the numerator and denominator of this expression respectively: $$= \frac{\cos^{2}{\frac{1}{2}x}}{ \sin{\frac{1}{2}x}\cos{\frac{1}{2}x}}$$ $$= \frac{\cos{\frac{1}{2}x}}{ \sin{\frac{1}{2}x}}$$ $$\cot{\frac{1}{2}x}$$, which is the right hand side. Hence, proved.

### Subject:Calculus

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Question:

Let $$y = (sin^{-1} x)^2$$ Prove that $$(1-x^{2})y^{\prime\prime} - xy^{\prime} = 2$$ Note: $$y^{\prime} and$$ $$y^{\prime\prime}$$ indicate the first derivative and second derivate of $$y$$ respectively.

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Nishant G.

The above question deals with proving that a differential equation for a function $$y$$ holds true, given the function. In calculus, it is often important to derive these differential equations from a function, so that they can be solved. The problem not only requires differentiation but also ingenuity in manipulating the function in order to prove the result. At the face value, it may seem difficult, but with some algebraic manipulation to the original function the result can be proved quite easily. The first instinct one might get is to just differentiate the original function, differentiate it again, and plug it in the expression to prove. However, this process might be tedious. Notice that the $$(1-x^{2})$$ term is there in the expression to prove. A starting point would be to multiply the original function by $$(1-x^{2})$$ in order to get: $$(1 - x^{2})y = (1-x^{2}) (sin^{-1} x)^2$$ Now, we have a starting point to differentiate both sides using the product rule to get: $$(1 - x^{2})y^{\prime} + y(-2x) = (1-x^{2})(\frac{1}{\sqrt{1 - x^{2}}}. 2sin^{-1} x) + (-2x) (sin^{-1} x)^2$$ Note that in order to get the above expression, we use the fact that $$\frac{d}{dx}sin^{-1} x = y^{\prime} = \frac{1}{\sqrt{1 - x^{2}}}$$ and apply normal differentiation with the product and chain rules. Since $$y = (sin^{-1} x)^2$$, the above expression we have so far can be simplified to: $$(1 - x^{2})y^{\prime} + y(-2x) = (1-x^{2})(\frac{1}{\sqrt{1 - x^{2}}}. 2\sqrt{y}) + (-2x)y$$ $$= (1 - x^{2})y^{\prime} = (1-x^{2})(\frac{1}{\sqrt{1 - x^{2}}}. 2\sqrt{y})$$ $$= (1 - x^{2})y^{\prime} = \sqrt{1-x^{2}}( 2\sqrt{y})$$ $$\implies ( \sqrt{1-x^{2}})y^{\prime} = 2\sqrt{y}$$. We will call this expression $$(a)$$. From $$(a)$$, we get an expression for $$y^{\prime} = \frac{2\sqrt{y}}{ \sqrt{1-x^{2}}}$$ . Since the expression we want to prove has a $$y^{\prime\prime}$$ term in it, we differentiate $$(a)$$ using the product rule again: $$\sqrt{1-x^{2}} y^{\prime\prime} + y^{\prime} \frac{-2x}{2\sqrt{1-x^2}} = \frac{2y^{\prime}}{2\sqrt{y}}$$ $$\implies \frac{ (1-x^{2}) y^{\prime\prime} - xy^{\prime} }{\sqrt{1-x^2}} = \frac{y^{\prime}}{\sqrt{y}}$$ $$\implies (1-x^{2}) y^{\prime\prime} - xy^{\prime} = \frac{y^{\prime}\sqrt{1-x^2}}{\sqrt{y}}$$ But, we know $$y^{\prime} = \frac{2\sqrt{y}}{ \sqrt{1-x^{2}}}$$$$\implies \frac{y^{\prime}\sqrt{1-x^2}}{\sqrt{y}} = \frac{2\sqrt{y}\sqrt{1-x^2}}{\sqrt{1-x^2}\sqrt{y}} = 2$$. Substituting this to our current expression, we have: $$\implies (1-x^{2}) y^{\prime\prime} - xy^{\prime} =2$$. Hence, proved. This problem made use of implicit differentiation, because we expressed the first derivative of $$y$$ in terms of $$y$$, in addition to some algebraic manipulation and normal differentiation (for example using the product rule and chain rules to differentiate the expressions) in order to finally prove the result.

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