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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.