Question

Please help me with the question provided in the attachment..

No spams please.. /(^_^) ​

Answer 1

Answer:

so, the option is b and c.

Step-by-step explanation:

sec²A  + sec²B = sec²A . sec²B

=> sec²A = sec²A . sec²B - sec²B

=>sec²A = sec²B (sec²A - 1)

    ( ∵ sec²x - 1 = tan²x)

=>sec²A = sec²B . tan²A

=>sec²A/tan²A = sec²B

=>(1/cos²A) / (sin²A/cos²A)

=>1/sin²A = sec²B

=>sin²A = cos²B

=>sin²A = sin²(90 - B)

=>sin²A - sin²(90 - B) = 0

=>(sinA + sin(90 - B))(sinA - sin(90-B)) = 0

∵ A and B are below 180° and in both quadrant sin is positive

so, => sinA - sin(90 - B) = 0

=>sinA = sin(90 - B)

so, => A = nπ + (-1)^n(π/2 - B)

 (i)  put n = 0

 => A = π/2 - B

 => A + B = π/2

(ii) put n = 1

=> A = π - (π/2 - B)

=> A = π/2 + B

(iii) put n = 2

=> A = 2π + (π/2 - B)

=> A + B = 5π/2

 This is not exist in Trangle ...

so, the Angle has right angle and obtuse angle.

Answer 2

Answer:

Correct option is (b) and (c).

Step-by-step explanation:

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