Question

In a right triangle ABC, right-angled at c, if tan A = 1, then verify that sec A cosec A=2.

Answer 1

Answer:

secA.cosecA = 2

Step-by-step explanation:

= > tanA = 1 ; tanA = 1

= > tanA = tan45° ; sinA / cosA = 1

= > A = 45° ; sinA = cosA

= > A = 45° ; 1 / sinA = 1 / cosA = secA = cosecA

Here,

= > secAcosecA

= > ( 1 / cosA )( 1 / sinA )

= > ( 1 / cosA )( 1 / cosA )

= > ( 1 / cos^2 A )

= > 1 / cos^2 45°

= > 1 / ( 1 /√2 )^2

= > 2

Hence,now, it's verified that in a right triangle ABC, right-angled at c, if tan A = 1, then verify that sec A cosec A=2.

Answer 2

Question :-- In a right triangle ABC, right-angled at c, if tan A = 1, then verify that sec A cosec A=2.

Formula used :--

→ tan45° = 1

→ sec45° = √2

→ cosec45° = √2

Solution :--

given, tan A = 1 ,

putting 1 = tan45° we get ,

→ tanA = tan45°

comparing now we get,

→ A = 45° ---------- Equation (1)

Now, we have to find , sec A cosec A = ?

→ sec A cosec A

Putting value of A From Equation (1) we get,

→ Sec45°* cosec45°

Putting values now we get,

→ (√2) * (√2)

→ 2

✪✪ Hence Proved ✪✪

So,

sec A cosec A=2.