Math, asked by AsifAhamed4, 1 year ago

please help me solve the 2nd question!

Don't spam

BEST ANSWER WILL BE MARKED AS BRAINLIEST

Attachments:

Answers

Answered by siddhartharao77
4

Note:

(1) sec²θ - tan²θ = 1 (or) sec²θ = 1 + tan²θ.

(2) cosec²θ - cot²θ = 1 (or) cosec²θ = 1 + cot²θ.

(3) cotθ = 1/tanθ.


Now,

Given : √sec²θ + cosec²θ

⇒ √(1 + tan²θ) + (1 + cot²θ)

⇒ √1 + tan²θ + 1 + cot²θ

⇒ √2 + tan²θ + cot²θ

⇒ √2 * tanθ * cotθ + tan²θ + cot²θ {From (3)}

⇒ √(tanθ + cotθ)²

⇒ tanθ + cotθ.


Hope it helps!


siddhartharao77: welcome
AsifAhamed4: can you help me any time?
AsifAhamed4: because I have a lot of maths doubt questions like this!
AsifAhamed4: I can give you even more points
siddhartharao77: i will try..
AsifAhamed4: hmmm..
AsifAhamed4: happy to hear that
AsifAhamed4: I have asked a new question
siddhartharao77: Come to inbox!
AsifAhamed4: I need more answers for coming to inbox
Similar questions