Math, asked by alpeshg398, 22 days ago

prove that : sin48°sec42°+cos48°cosec42°=2

Answers

Answered by ItzSofiya

Step-by-step explanation:

Given sin48

∘

sec42

∘

+cos48

∘

csc42

∘

we know that,

secθ=

cosθ

and cscθ=

sinθ

∴

cos42

∘

sin48

∘

sin42

∘

cos48

∘

We have,

cos(90−θ)=sinθ, sin(90−θ)=cosθ

∴

cos(90−48)

∘

sin48

∘

sin(90−48)

∘

cos48

∘

sin48

∘

sin48

∘

cos48

∘

cos48

∘

sin48

∘

cos48

∘

cos48

∘

=1+1

∴ sin48

∘

sec42

∘

+cos48

∘

csc42

∘

hence proved

Answered by heijsk1

Given sin48
sec42
+cos48
csc42
we know that,
secθ=
cosθ
1
and cscθ=
sinθ
1
∴
cos42
sin48
+
sin42
cos48
We have,
cos(90−θ)=sinθ, sin(90−θ)=cosθ
∴
cos(90−48)
sin48
+
sin(90−48)
cos48
=
sin48
sin48
+
cos48
cos48
=1+1
=2
∴ sin48
sec42
+cos48
csc42
=2

hence proved

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