Question

(SecA+ cos A) (secA - cos A) ​

Answer 1

• $\orange{\textsf{(Sec A + Cos A) (Sec A - Cos A)}}$

We know that,

• (a - b) (a + b) = a² - b²

• Here,

• a = Sec A

• b = Cos A

Substituting,

• Sec² A - Cos ² A

We know that,

• 1 + Tan² A = Sec² A

• 1 - Sin² A = Cos² A

Substituting,

• 1 + Tan² A - (1 - Sin² A)

• 1 + Tan² A - 1 + Sin² A

• Tan² A + Sin² A

Therefore,

• $\boxed{\purple{\textsf{(Sec A + Cos A) (Sec A - Cos A) = Tan² A + Sin² A}}}$

Answer 2

To find:-

(SecA+ cos A) (secA - cos A)

Formulae used:-

• (a - b) (a + b) = a² - b²
• 1 + Tan² A = Sec² A
• 1 - Sin² A = Cos² A

And here,

a = Sec A

b = Cos A

Solution:-

By substituting the values,

$= 1 + Tan² A - (1 - Sin² A)$

$= 1 + Tan² A - 1 + Sin² A$

$= Tan² A + Sin² A$

$\mathtt{therefore,(SecA+ cosA)(secA - cos A)}$

$\mathtt{= Tan² A + Sin² A\: }$