# Tutor profile: Paresh D.

## Questions

### Subject: Trigonometry

Prove the following identity: (1 - sin A)/(1 + sin A) = (sec A - tan A)2

In order to solve such questions, we have to first of revise all the important trigonometric formulas and use tricks to solve the problem in a minimum amount of steps. Now, how can one learn these tricks? Well, the answer is through practice. The more you practice, the more you learn to solve a question fast. So, let u begin with this problem, divinding it into two parts, i.e. L.H.S.(Left Hand Side) and R.H.S.(Right Hand Side). Solving the L.H.S. first: L.H.S = (1 - sin A)/(1 + sin A) = (1 - sin A)2/(1 - sin A) (1 + sin A), [Multiply both numerator and denominator by (1 - sin A)] This trick came to my mind because sin2A+cos2A=1, and we can substitute the 1-sin2A term as cos2A. = (1 - sin A)2/(1 - sin2 A) = (1 - sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 - sin2 θ] = {(1 - sin A)/cos A}2 = (1/cos A - sin A/cos A)2 = (sec A – tan A)2 = R.H.S. Hence Proved.

### Subject: Calculus

Let the closed interval [a , b] be the domain of function f(x). What is the domain of f(x - 3)?

Domains, graphs and analysis questions are often the type of questions students get confused with. The secret to solving this kind of questions is understanding what is actually happening and then arriving at the results. Now, let us take up this question. Domain of a function is given and it is asket if the function changes from f(x) to f(x-3) then what will happen to the domain. Now, understand what is happening. If you put different values of x in this equation, the input that the function gets is always x-3. This means that the domain, the values that can be inputted in the function is reduced by 3. Or in other words, The graph of f(x - 3) is that of f(x) shifted 3 units to the right. Now, in order to find the new domain of this function, we have to shift the end points of this domain by 3 units to the right and to do this we add 3 units to both the endpoints. So, the new domain of f(x-3) becomes [a+3, b+3]

### Subject: Algebra

If m and p are the roots of the equation Ax2-4x+1 = 0 and n and q are the roots of Bx2-6x+1 = 0. find the value of B-A such that m,n,p,q are in H.P.(Harmonic Progression).

Okay, so they have given us two equations, the roots of which form an H.P. and we have to find the difference between the coefficients of the equations. So, to solve this question, we should first know what an H.P. is. So, let us revise what is an H.P. in context with the problem. H.P. can very well be understood by an A.P. So, if m,n,p,q are in H.P., 1/m,1/n,1/p,1/q are in A.P. Okay, if this is clear, let us move forward. m+p=4/A and m*p=1/A Combining these 2 equations, we get: (Substitute the value of 1/A as m*p) 1/m+1/p=4 equation no. 1 n+q=6/B and n*q=1/B Combining these 2 equations, we get: (Substitute the value of 1/b as n*q) 1/n+1/q=6 equation no. 2 Now, let us assume that there is a common difference 'd' in the A.P. formed by 1/m,1/n,1/p,1/q. So, the new equations will be:(From equation no. 1 and 2) 2/m+2*d=4 (equation 3) and 2/m +4*d=6 (equation 4) Subtracting the above equations, we obtain: 2*d=2 d=1 Substituting this value of d in equation 3, we get: m=1. Now, 1/p=1/m+2*d 1/p=1+2 p=1/3 (Equation 5) Also, 1/n=1/m+d 1/n=1+1 n=1/2 (Equation 6) 1/q=1/m+3*d 1/q=1+3 q=1/4 (Equation 7) Now, from Equation 5,6,7, we get: 1/A=m*p=1/3, A=3 1/B=n*q=1/8, B=8 Therefore, we get B-A=8-3=5 Final Answer

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