Question

Answer this question........
question 24 ,27

Answer 1

27>

sec^2 θ+ cosec^2 θ

=1/cos^2 θ + 1/sin^2 θ

=(sin^2 θ + cos^2 θ) / cos ^2 θ sin^2 θ

=1/cos^2 θ sin^2 θ

=sec^2 θ cosec^2 θ (proved)  

 

Answer 2

⭐⭐⭐⭐⭐ ANSWER ⭐⭐⭐⭐⭐



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24) If we hold "theta" as "x", then

tan x + sin x = a
tan x - sin x = b

So, ab = (tan x + sin x) (tan x - sin x)
=> ab = tan²x - sin²x
=> ab = sin²x/cos²x - sin²x
=> ab = sin²x (1/cos²x - 1)
=> ab = sin²x (sec²x-1)
=> ab = sin²x tan²x
=> sin x tan x = √ab.......... (i)


Again, tan x = 1/2(a+b)
sin x = 1/2(a-b)

So, tan x sin x = 1/4(a²-b²) .............. (ii)

From (i) and (ii), we get,

√ab = 1/4(a²-b²)
=> 4√ab = (a²-b²) [PROVED]

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27) Let the theta be "x".

So, cosec²x + sec²x
= 1/sin²x + 1/cos²x
= cos²x + sin²x / cos²x sin²x
= 1/cos²x sin²x [Since, cos²x + sin²x = 1]
= 1/cos²x × 1/sin²x
= cosec²x sec²x [PROVED]

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⭐⭐⭐ ALWAYS BE BRAINLY ⭐⭐⭐

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