How do you verify (sec θ + csc θ)(cos θ - sin θ) = cot θ - tan θ?

Solution:

Given, (sec θ + csc θ)(cos θ - sin θ) = cot θ - tan θ

Consider L.H.S.,

(sec θ + cosec θ)(cos θ - sin θ)

= [(1 / cos θ) + (1 / sin θ)] (cos θ - sin θ)

= [(sin θ + cos θ) / (sin θ. cos θ)] (cos θ - sin θ)

= (cos² θ - sin² θ) / (sin θ. cos θ)

= [(cos² θ / (sin θ. cos θ)] - [sin² θ / (sin θ. cos θ)]

= (cos θ / sin θ) - (sin θ / cos θ)

= cot θ - tan θ 

L.H.S. = R.H.S

Aliter

We can also verify by taking θ with a particular angle.

Let θ = 60°, 

Now, considering L.H.S. 

L.H.S. = (sec θ + cosec θ)(cos θ - sin θ)

= (sec 60° + cosec 60°) (cos 60° - sin 60°)

= [2 + (2/√3)] [(1/2) - (√3/2)]

= [2(1 + 1/√3)] [(1 - √3)/2]

= [1 + (1/√3)] (1 - √3)

= [(√3 + 1)/ √3] (1 - √3)

= (1 - 3) / √3

= -2/ √3 -----(1)

R.H.S. = cot θ - tan θ

= cot 60° - tan 60°

= (1/√3) - √3

= (1 - 3) / √3

= -2/√3 ------(2)

From (1) and (2) 

(sec θ + csc θ)(cos θ - sin θ) = cot θ - tan θ

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How do you verify (sec θ + csc θ)(cos θ - sin θ) = cot θ - tan θ?

Summary:

We can verify the equation (sec θ + csc θ)(cos θ - sin θ) = cot θ - tan θ by solving simultaneously LHS and RHS which is done in method I or by assuming the values of ‘θ’as done in method II.