Ex 8.3

Ex 8.3, 1

Ex 8.3, 2 Important

Ex 8.3, 3 (i) [MCQ]

Ex 8.3, 3 (ii) [MCQ] Important

Ex 8.3, 3 (iii) [MCQ] Important

Ex 8.3, 3 (iv) [MCQ]

Ex 8.3, 4 (i) Important

Ex 8.3, 4 (ii)

Ex 8.3, 4 (iii) Important You are here

Ex 8.3, 4 (iv) Important

Ex 8.3, 4 (v) Important

Ex 8.3, 4 (vi)

Ex 8.3, 4 (vii) Important

Ex 8.3, 4 (viii)

Ex 8.3, 4 (ix) Important

Ex 8.3, 4 (x)

Question 1 (i) Important Deleted for CBSE Board 2025 Exams

Question 1 (ii) Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii)tanθ/(〖1 − cot〗θ " " )+cotθ/(1 − tanθ ) =1+ sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] Taking L.H.S tanθ/(〖1 − cot〗θ " " )+cotθ/(1 − tanθ ) Converting everything into sin θ and cos θ = (sinθ/cosθ )/(1 − (cosθ/sinθ ) )+((cosθ/sinθ ))/(1 − (sinθ/cosθ ) ) = (sinθ/cosθ )/(((sinθ − cosθ)/sinθ ) )+((cosθ/sinθ ))/( ((cosθ − sinθ)/cosθ ) ) = sinθ/cosθ × sinθ/(sinθ −〖 cos〗θ ) +cosθ/sinθ ×cosθ/(cosθ −〖 sin〗θ ) = sin2θ/(〖cosθ ( sin〗θ −〖 cos〗〖θ )〗 ) + cos2θ/(sin〖θ 〗〖(𝐜𝐨𝐬𝜽 〗 −〖 𝒔𝒊𝒏〗〖𝜽 )〗 ) = sin2θ/(〖cosθ ( sin〗θ −〖 cos〗〖θ )〗 ) + cos2θ/(〖sin〖θ ×〗 − ( 𝐬𝐢𝐧〗𝜽 −〖 𝒄𝒐𝒔〗〖𝜽 )〗 ) = sin2θ/(〖cosθ (sin〗θ −〖 cos〗〖θ )〗 ) − cos2θ/(〖sinθ ( sin〗θ − cos〖θ )〗 ) = (𝑠𝑖𝑛2𝜃 × (𝑠𝑖𝑛𝜃 ) −〖 𝑐𝑜𝑠2〗𝜃 × (𝑐𝑜𝑠𝜃 ))/(〖𝑐𝑜𝑠𝜃 ( 𝑠𝑖𝑛〗𝜃 −〖 𝑐𝑜𝑠〗〖𝜃 )𝑠𝑖𝑛𝜃 〗 ) = (𝐬𝐢𝐧𝟑𝜽 − 𝐜𝐨𝐬𝟑𝜽 )/(〖cosθ sinθ ( sin〗θ −〖 cos〗〖θ)〗 ) Since a3 – b3 = (a – b) (a2 + b2 + ab) Putting a = sin θ , b = cos θ = (〖(sin〗θ − cosθ)〖(sin2〗θ+ cos2〖θ +cosθ sinθ 〗))/(〖cosθ sinθ ( sin〗θ −〖 cos〗〖θ ) 〗 ) = (〖(sin2〗θ + cos2〖θ + cosθ sinθ 〗))/(cosθ sinθ ) = (〖(𝐬𝐢𝐧𝟐〗𝜽 + 𝐜𝐨𝐬𝟐〖𝜽 ) +〖 cos〗θ sinθ 〗)/(cosθ sinθ ) As cos2 A + sin2 A = 1 = (𝟏 + cosθ sinθ)/(cosθ sinθ ) = (1 )/(cosθ sinθ ) + (cosθ sinθ)/(cosθ sinθ ) = (𝟏 )/𝒄𝒐𝒔𝜽 × (𝟏 )/𝒔𝒊𝒏𝜽 + 1 = sec θ × cosec θ + 1 = 1 + sec θ cosec θ = R.H.S ∴ L.H.S = R.H.S Hence proved