Question

Evaluate 8/cot square theta -8 /cos square theta

Answer 1

8/cot²A - 8/cos²A = - 8

Trigonometry: Trigonometry is the study of angles and relations between angles and their sin, cos, tan, cosec, sec, cot ratios. There are many formulae for calculations:

• sin²A + cos²A = 1

• sec²A - tan²A = 1

• cosec²A - cot²A = 1

• sin2A = 2 sinA cosA

• cos2A = cos²A - sin²A

• tan2A = 2 tanA / (1 - tan²A)

• cotA = cosA / sinA

• tanA cotA = 1

Solution:

Method 1.

Now, 8/cot²A - 8/cos²A

= 8 (1/cot²A - 1/cos²A)

= 8 {1/(cos²A/sin²A) - 1/cos²A}

= 8 (sin²A/cos²A - 1/cos²A)

= 8 * (sin²A - 1)/cos²A

= - 8 * (1 - sin²A)/cos²A

= - 8 * cos²A/cos²A

= - 8 * 1

= - 8

Method 2.

Now, 8/cot²A - 8/cos²A

= 8 (1/cot²A - 1/cos²A)

= 8 (tan²A - sec²A)

= - 8 (sec²A - tan²A)

= - 8 * 1, [ since sec²θ - tan²θ = 1 ]

= - 8

Answer 2

$\frac{8}{cot^{2}\alpha}-\frac{8}{cos^{2}\alpha}=-8$

Solution:

  $\frac{8}{cot^{2}\alpha}-\frac{8}{cos^{2}\alpha}=-8(\frac{1}{cos^{2}\alpha}-\frac{1}{cot^{2}\alpha})$

  $\frac{8}{cot^{2}\alpha}-\frac{8}{cos^{2}\alpha}=-8({sec^{2}\alpha}-{tan^{2}\alpha})$

                                       [∵ $\frac{1}{cos^{2}\alpha}=sec^{2}\alpha$

                                           $\frac{1}{cot^{2}\alpha}=tan^{2}\alpha$]

   $\frac{8}{cot^{2}\alpha}-\frac{8}{cos^{2}\alpha}=-8(1)$

                                        [∵ $sec^{2}\alpha}-{tan^{2}\alpha=1$]

   $\frac{8}{cot^{2}\alpha}-\frac{8}{cos^{2}\alpha}=-8$