Question

Express cosec theta - cot theta whole square interms of cos

Answer 1

AnswEr

$\frac{1 - \cos \theta}{1 + \cos \theta}$

Solution

Given ,

$( { \cosec \theta - \cot \theta})^{2} \\ \\ = (\frac{1}{ \sin \theta} - \frac{ \cos \theta}{ \sin \theta} ) ^{2} \\ \\ = ( \frac{1 - \cos \theta}{ \sin \theta} ) ^{2} \\ \\ = \frac{( {1 - \cos \theta)}^{2} }{ { \sin}^{2} \theta} \\ \\ = \frac{ {(1 - \cos \theta)}^{2} }{1 - { \cos}^{2} \theta} \\ \\ = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \\ \\ = \frac{1 - \cos \theta}{1 + \cos \theta}$

Formula used here

• cosecA = 1/sinA

• cotA = cosA/sinA

• (a² - b²) = (a + b)(a - b)

• sin²A + cos²A = 1

Other formulae related to it

• secA = 1/cosA

• tanA = sinA/cosA

• tan²A + 1 = sec²A

• cot² + 1 = cosec²A

• sin(A + B) = sinAcosB + cosAsinB

• sin(A - B) = sinAcosB - cosAsinB

•cos(A + B) = cosAcosB - sinAsinB

• cos(A - B) = cosAcosB + sinAsinB

•tan(A + B)=(tanA+tanB)/(1 -tanAtanB)

•tan(A - B) =(tanA -tanB/(1 +tanAtanB)

•sin2A = 2sinAcosA

• cos2A = cos²A - sin²A= 2cos²A -1

= 1 - 2sin²A

• tan2A = 2tanA/(1+tan²A)