Question 14 - NCERT Exemplar - MCQ - Chapter 8 Class 10 Introduction to Trignometry (Term 1)
Last updated at Oct. 20, 2021 by

The value of the expression [(sin
^{
2
}
22
^{
°
}
+ sin
^{
2
}
68
^{
°
}
)/(cos
^{
2
}
22
^{
°
}
+ cos
^{
2
}
68
^{
°
}
) + sin
^{
2
}
63
^{
°
}
+ cos 63
^{
°
}
sin27
^{
°
}
] is
(A) 3 (B) 2 (C) 1 (D) 0

Transcript

Question 14
The value of the expression [(sin^222"°" + sin^268"°" )/(cos^2〖22"°" 〗+ cos^268"°" )+sin^2〖63"°" +cos〖63"°" sin27"°" 〗 〗 ] is
(A) 3 (B) 2 (C) 1 (D) 0
(sin^222"°" + sin^268"°" )/(cos^2〖22"°" 〗+ cos^268"°" )+sin^2〖63"°" +cos63"°" 𝒔𝒊𝒏𝟐𝟕"°" 〗
Using sin θ = cos (90° − θ)
= (sin^222"°" + sin^268"°" )/(cos^2〖22"°" 〗+ cos^268"°" )+sin^2〖63"°" +cos〖63"°" 𝐜𝐨𝐬〖(𝟗𝟎°−𝟐𝟕"°)" 〗 〗 〗
= (sin^222"°" + sin^268"°" )/(cos^2〖22"°" 〗+ cos^268"°" )+sin^2〖63"°" +cos〖63"°" 𝒄𝒐𝒔〖𝟔𝟑°〗 〗 〗
= (sin^222"°" + sin^268"°" )/(cos^2〖22"°" 〗+ cos^268"°" )+〖𝒔𝒊𝒏〗^𝟐〖𝟔𝟑"°" +〖𝒄𝒐𝒔〗^𝟐𝟔𝟑"°" 〗
Using cos2 θ + sin2 θ = 1
(sin^222"°" + 〖𝒔𝒊𝒏〗^𝟐𝟔𝟖"°" )/(cos^2〖22"°" 〗+ 〖𝒄𝒐𝒔〗^𝟐𝟔𝟖"°" )+1
= (sin^222"°" + 〖𝒄𝒐𝒔〗^𝟐(𝟗𝟎° − 𝟔𝟖"°" ))/(cos^2〖22"°" 〗+ 〖𝒔𝒊𝒏〗^𝟐(𝟗𝟎° − 𝟔𝟖"°" ) )+1
= (sin^222"°" + 〖𝒄𝒐𝒔〗^𝟐𝟐𝟐"°" )/(cos^2〖22"°" 〗+ 〖𝒄𝒐𝒔〗^𝟐𝟐𝟐"°" )+1
= 1+1
= 2
So, the correct answer is (B)

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