Trigonometry questions

Question

Q1 : If 7 sin² Ø + 3 cos² Ø = 4, and  0º ≤ Ø ≤ 90º, find the value of sec Ø + cosec Ø.

 

Q2 : If x = a secØ + b tan Ø, y = a tan Ø + b sec Ø, then prove that x² - y² = a² - b^2.

 

Q3 : If cot B = 12/5, prove that tan² B - sin² B = sin⁴ B . sec² B

 

 

Answer 1

Let    x=Ø  for all questions.

 

 

Q1)

 

Get the equation in terms of either sine of cosine.

 

7sin²x + 3(1 - sin²x) = 4

 

7sin²x - 3sin²x + 3 = 4

 

4sin²x - 1 = 0

 

4sin²x = 1

 

  sin²x = 1/4

 

  sinx = ±1/2

 

 

Since the angle is the first quadrant.

 

sinx = 1/2

 

 

Then we find cosine using the substitution.

 

cosx = √(1 - (1/2)²)

cosx = √(1 - (1/4))

cosx = √(3/16)

cosx = √(3) / 4

 

Then   secx + cscx = (1 / cosx) + (1 / sinx)

 

Plug in the obtained values of cosx and sinx into this expression.

 

 

Q2)

 

x² - y² = (x - y)(x + y)

 

          = [(asecx + btanx) - (atanx + bsecx)] * [(asecx + btanx) + (atanx + bsecx)]

 

 

Combine like terms and factor where possible.

 

          = [secx(a - b) - tanx(a - b)] * [secx(a + b) + tanx(a + b)]

 

 

Next, look for the common factor in each piece.

 

          = [(a - b)(secx - tanx)] * [(a + b)(secx + tanx)]

 

          = (a - b)(a + b)(sec²x - tan²x)

 

 

Using the identity, we get

 

          = (a - b)(a + b)[(1 / cos²x) - (sin²x / cos²x)]

 

          = (a - b)(a + b)[cos²x / cos²x)]

 

          = (a - b)(a + b)

 

          = a² - b²

 

 

Now try the last one on your own using the identity for cotangent.  It is similar to Q2.