Divakar B.

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Trigonometry

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Question:

Prove the following identity: 1+2sec^2Atan^2A-sec^4A-tan^4A = 0

Divakar B.

Answer:

Solution: LHS = 1+2sec^2Atan^2A-sec^4A-tan^4A = 1 - (sec^4A - 2sec^2A tan^2A + tan^4A) Now we can write value inside brackets as (a-b)^2 form where a=sec^2A and b=tan^2A = 1 - ((sec^2A)^2 - 2sec^2A tan^2A + (tan^2A)^2) = 1 - (sec^2A - tan^2A)^2 Now use the identity sec^2 A - tan^2 A = 1 = 1-1 = 0 = RHS Hence Proved

Calculus

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Question:

Differentiate y=sin(6x)

Divakar B.

Answer:

The outer layer is "the sine function" and the inner layer is (6x) . Differentiate "the sine function" first, leaving (6x) unchanged. Then differentiate (6x). D(sin(6x)) = cos(6x) D(6x) Now derivative of 6x is 6 So.. D(sin(6x)) = cos(6x) (6) D(sin(6x)) = 6cos(6x) So result is: 6cos(6x)

Algebra

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Question:

Solve sqrt(x)+sqrt(x-7) = 1

Divakar B.

Answer:

First make a note of the fact that you cannot take the square root of a negative number. Therefore,sqrt(x) term is valid only if sqrt(x) >= 0 and the second term sqrt(x-7) is valid if x-7 >=0..... x >=7 Isolate sqrt(x-7) term..for that subtract sqrt(x) on both sides sqrt(x)+sqrt(x-7)-sqrt(x) = 1-sqrt(x) sqrt(x-7) = 1-sqrt(x) Square both sides of the equation. (sqrt(x-7))^2 = (1-sqrt(x))^2 x-7 = 1-2sqrt(x)+x Subtract x on both sides x-7-x = 1-2sqrt(x)+x-x -7 = 1-2sqrt(x) 2sqrt(x)=7+1 2sqrt(x)=8 sqrt(x)=4 Apply square on both sides (sqrt(x))^2 = 4^2 x = 16 Check the solution by substituting 16 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer. Left side (Plug x=16) : sqrt(16)+sqrt(16-7) = 4+3 = 7 Right side : 1 Since the left side of the original equation does not equal the right side of the original equation after we substituted our solution for x, then there is no solution. So result is : NO SOLUTION

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