EQUATION OF TANGENT AND NORMAL TO THE CURVE AT THE GIVEN POINT

Find the equation of the tangent and normal of the following curves

(i)  y  =  x²-4x-5, at x = -2

(ii)  y  =  x-sin x cos x, at x = π/2

(iii)  y  =  2sin²3x, at x = π/6

(iv)  y  =  (1+sin x)/cos x, at x  =  π/4

Question 1 :

y  =  x²-4x-5, at x = -2

When x  =  -2, then

So, the required point is (-2, 7)

Question 2 :

y  =  x-sin x cos x, at x = π/2

Solution :

y  =  x-sin x cos x when x = π/2.

dy/dx  =  1-[sin x (- sin x)+cos x (cos x)]

=  1-[- sin²x + cos²x]

=  1-[cos²x-sin²x]

=  1-cos 2x

Slope of the tangent at x  =  π/2

=  1-cos2(π/2)

=  1-cos π

=  2

So, the required point is (π/2, π/2)

Question 3 :

y  =  2sin²3x, at x = π/6

Solution :

y  =  2sin² 3x

dy/dx  =  2(2sin3x) (cos3x) 3

=  6(2sin3x cos3x)  

=  6sin 2(3x)

=  6sin6x

So, the required point is (π/6,2)

Question 4 :

y  =  (1+sin x)/cos x, at x  =  π/4

Solution :

y  =  (1+sin x)/cos x

dy/dx  =  [cos x (cos x) - (1 + sin x) (-sin x)]/cos² x

=  [cos²x+sinx+sin²x]/cos²x

=  (1+sinx)/cos²x

Slope at x  =  π/4

y  =  (1+sinπ/4)/cos²π/4

=  (1+(1/√2))/(1/√2)²

=  [(√2+1)/√2]/(1/2)

=  [(√2+1)/√2] ⋅ (2/1)

=  (√2+1)√2

y  =  2+2√2

y  =  (1+sin x)/cos x

=  (1+sinπ/4)/cosπ/4

=  [1+(1/√2)]/(1/√2)

=  [(√2+1)/√2]/(√2/1)

=  (√2+1)

So, the required point is (π/4, (√2+1))

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