Question

√secA-tanA/secA+tanA × √secA+cotA/secA+cotA= Sec A - tan A + cosec A- cot A​

Answer 1

$\huge \red{question}$✍

Solution : ⤵️

Start by expanding out to primitive trig functions. The left-hand side expands as:

secA+tanAcscA+cotA=1+sinAcosA1+cosAsinA=1+sinA1+cosAtanAsec⁡A+tan⁡Acsc⁡A+cot⁡A=1+sin⁡Acos⁡A1+cos⁡Asin⁡A=1+sin⁡A1+cos⁡Atan⁡A

The right-hand side expands as:

cscA−cotAsecA−tanA=1−cosAsinA1−sinAcosA=1−cosA1−sinAcotAcsc⁡A−cot⁡Asec⁡A−tan⁡A=1−cos⁡Asin⁡A1−sin⁡Acos⁡A=1−cos⁡A1−sin⁡Acot⁡A

Our objective is to multiply both sides by the same factors,

simplifying them piece by piece until they are obviously equal.

To start let’s multiply the expanded left-hand side and right-hand side by tanAtan⁡A.

The left-hand side becomes:

1+sinA1+cosAtan2A1+sin⁡A1+cos⁡Atan2⁡A

and the right-hand side becomes:

1−cosA1−sinA1−cos⁡A1−sin⁡A

That 1−sinA1−sin⁡A screams at us to use the difference of two squares.

Let’s multiply both sides by 11+sinA11+sin⁡A.

This will also cancel out a term on the left-hand side.

The left hand side becomes:

1−cos2Acos2A=sin2Acos2A=tan2A1−cos2⁡Acos2⁡A=sin2⁡Acos2⁡A=tan2⁡A

And of course tan2A=tan2Atan2⁡A=tan2⁡A. Notice that we must have sinA≠0,cosA≠0,cosA≠−1,sinA≠1sin⁡A≠0,cos⁡A≠0,cos⁡A≠−1,sin⁡A≠1. So A≠kπ2A≠kπ2 for integer k

tan2A1+cosAtan2⁡A1+cos⁡A

The right-hand side becomes: