Question 19
Evaluate: (cosec^2〖63°〗 + tan^2〖24°〗)/(cot^2〖66°〗 + sec^2〖27°〗 ) + (sin^2〖63°〗 + cos〖63°〗 sin〖27°〗 + sin〖27°〗 sec〖63°〗)/(2(cosec^2〖65°〗 − tan^2〖25°〗))
(cosec^2〖63°〗 + tan^2〖24°〗)/(cot^2〖66°〗 + sec^2〖27°〗 ) + (sin^2〖63°〗 + cos〖63°〗 sin〖27°〗 + sin〖27°〗 sec〖63°〗)/(2(cosec^2〖65°〗 − tan^2〖25°〗))
Theory
We know that we have to use
sin (90 – θ) = cos θ
sin becomes cos
cos becomes sin
tan becomes cot
cot becomes tan
cosec becomes sec
sec becomes cosec
So, we need to form pairs
Theory
This will become
sin2 63° + cos2 63°
And we can use
sin2 x + cos2 x = 1
This will become
sin 27° × cosec 27°
And we put
cosec x = 1/sin𝑥
(〖𝒄𝒐𝒔𝒆𝒄〗^𝟐〖𝟔𝟑°〗 + 〖𝒕𝒂𝒏〗^𝟐〖𝟐𝟒°〗)/(〖𝒄𝒐𝒕〗^𝟐〖𝟔𝟔°〗 + 〖𝒔𝒆𝒄〗^𝟐〖𝟐𝟕°〗 ) + (〖𝒔𝒊𝒏〗^𝟐〖𝟔𝟑°〗 + 𝒄𝒐𝒔〖𝟔𝟑°〗 𝐬𝐢𝐧〖𝟐𝟕°〗 + 𝒔𝒊𝒏〖𝟐𝟕°〗 𝒔𝒆𝒄〖𝟔𝟑°〗)/(𝟐(〖𝒄𝒐𝒔𝒆𝒄〗^𝟐〖𝟔𝟓°〗 − 〖𝒕𝒂𝒏〗^𝟐〖𝟐𝟓°〗))
Denominator will become
tan^2〖24°〗+cosec^2〖63°〗
and get cancelled by numerator
Denominator will become
cosec^2〖65°〗 − cot^2〖65°〗
And we can use
cosec2 x – cot2 x = 1
(cosec^2〖63°〗 + tan^2〖24°〗)/(cot^2〖66°〗 + sec^2〖27°〗 ) + (sin^2〖63°〗 + cos〖63°〗 sin〖27°〗 + sin〖27°〗 sec〖63°〗)/(2(cosec^2〖65°〗 − tan^2〖25°〗))
Using cos θ = sin (90 – θ)
sin θ = cos (90 – θ)
tan θ = cot (90 – θ)
cot θ = tan (90 – θ)
sec θ = cosec (90 – θ)
cosec θ = sec (90 – θ)
= (cosec^2〖63°〗 + tan^2〖24°〗)/(tan^2〖(90° − 66°)〗 + cosec^2〖(90° − 27°〗)) + (sin^2〖63°〗 + cos〖63°〗 〖cos 〗〖(90° −27°)〗+ sin〖27°〗 cosec〖 (90° − 63°)〗)/2(cosec^2〖65°〗 −〖 cot〗^2〖(90° − 25°〗))
= (cosec^2〖63°〗 + tan^2〖24°〗)/(tan^2〖(24°)〗 + cosec^2〖(63°〗)) + (sin^2〖63°〗 + cos〖63°〗 cos〖 (63°)〗+ sin〖27°〗 cosec〖(27°)〗)/2(cosec^2〖65°〗 −〖 cot〗^2〖(65°〗))
= 1 + (sin^2〖63°〗 + cos^2〖63°〗+ sin〖27°〗 cosec〖27°〗)/2(cosec^2〖65°〗 −〖 cot〗^2〖65°〗 )
Using cosec2 x – cot2 x = 1
= 1 + (〖𝒔𝒊𝒏〗^𝟐〖𝟔𝟑°〗 + 〖𝒄𝒐𝒔〗^𝟐〖𝟔𝟑°〗+ sin〖27°〗 cosec〖27°〗)/2(1)
Using sin2 x + cos2 x = 1
= 1 + (1 + 𝑠𝑖𝑛〖27°〗 𝑐𝑜𝑠𝑒𝑐〖27°〗)/2
Using cosec x = 1/sin𝑥
= 1 + (1 + 𝑠𝑖𝑛〖27°〗 × 1/sin〖27°〗 )/2
= 1 + (1 +1)/2
= 1 + 2/2
= 1 + 1
= 2
Question 19
If sin 𝜃 + cos 𝜃 = √2, then evaluate : tan 𝜃 + cot 𝜃.
Given
sin θ + cos θ = √2
We need to find
tan θ + cot θ
Now,
tan θ + cot θ = sin𝜃/cos𝜃 +cos𝜃/sin𝜃
= (sin𝜃 × sin𝜃 + cos𝜃 × cos𝜃)/(cos𝜃 × sin𝜃 )
= (sin^2𝜃 + cos^2𝜃)/(cos𝜃 sin𝜃 )
Putting sin2 x + cos2 x = 1
= 1/(cos𝜃 sin𝜃 )
Thus,
tan θ + cot θ = 1/(cos𝜃 sin𝜃 )
Now
sin θ + cos θ = √2
Squaring both sides
(sin θ + cos θ)2 = (√2)^2
sin2 θ + cos2 θ + 2 sin θ cos θ = 2
Putting sin2 x + cos2 x = 1
1 + 2 sin θ cos θ = 2
2 sin θ cos θ = 2 – 1
2 sin θ cos θ = 1
sin θ cos θ = 1/2
From (1)
tan θ + cot θ = 1/(cos𝜃 sin𝜃 )
Putting value from (2)
tan θ + cot θ = 1/((1/2) )
tan θ + cot θ = 2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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