# Tutor profile: Shubham S.

## Questions

### Subject: Trigonometry

Express in terms of sin2A tanA + cotA

to find the answer in terms of sin2A we shall first write tan A and cotA in terms of sinA and cosA sinA/cosA + cosA/sinA=> ( $$ sin^{2}A + cos^{2}A $$ )/ sinAcosA => 1/sinAcosA, now multiply both numerator and denominator by 2 to get => 2/2sinAcosA =>2/sin2A as 2sinAcosA=sin2A Ans- 2/sin2A

### Subject: Basic Chemistry

HCl + NaOH -> ???

Since HCl is a strong acid and NaOH is a strong base, they mix to give a salt which is neutral and water so, HCl + NaOH -> NaCl + $$ H_{2}O $$ where NaCl is the neutral salt Note that this reaction gives a NEUTRAL salt only because both the acid and the base are STRONG

### Subject: Algebra

Given the two linear equations. Solve for x and y. 2x+3y=7, 7x+4y=13

From the two given equations, to solve for x and y we try and solve the 2 equations by eliminating 1 variable which can be done by making the coefficients of the variables in the equations same and then subtracting 1 from the other. (2x+3y=7) * 4=> 8x+12y=28 (7x+4y=13) * 3=>21x+12y=39 Now subtract the first equation from the second to get 13x=11 which gives us, x=11/13 now put this value of x in any one of the equation to get the value of y 2(11/13)+3y=7 => 3y=7-22/13 => 3y=69/13 => y=23/13 so x=11/13, y=23/13 is the solution of the given pair of linear equations.

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