Prove the following that-
tan 3 ⁡θ/1 + tan 2 ⁡θ + cot 3 ⁡θ/1 + cot 2 ⁡θ = sec θ cosec θ – 2 sin θ cos θ



CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard
CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard
Last updated at Dec. 13, 2024 by Teachoo
Transcript
Question 29 Prove the following that- tan^3θ/(1 + 〖𝑡𝑎𝑛^2〗θ ) + cot^3θ/(1 + 〖𝑐𝑜𝑡〗^2θ ) = sec θ cosec θ – 2 sin θ cos θSolving LHS tan^3θ/(1 + 〖𝑡𝑎𝑛^2〗θ ) + cot^3θ/(1 + 〖𝑐𝑜𝑡〗^2θ ) = tan^3θ/(1 + 〖𝑡𝑎𝑛^2〗θ ) + (𝟏/〖𝐭𝐚𝐧〗^𝟑𝜽 )/(𝟏 + 𝟏/〖𝐭𝐚𝐧〗^𝟐𝜽 ) = tan^3θ/(1 + 〖𝑡𝑎𝑛^2〗θ ) + (1/tan^3θ )/((tan^2θ + 1)/tan^2θ ) = tan^3θ/(1 + 〖𝑡𝑎𝑛^2〗θ ) + 1/tan^3θ ×tan^2θ/(tan^2θ + 1) = tan^3θ/(1 + 〖𝑡𝑎𝑛^2〗θ ) + 1/𝑡𝑎𝑛θ ×1/(tan^2θ + 1) = (tan^3θ ×tan〖θ 〗 + 1)/(tanθ (1 + 〖𝑡𝑎𝑛^2〗θ)) = (tan^4θ + 1)/(tanθ (𝟏 + 〖𝒕𝒂𝒏^𝟐〗𝜽)) Putting 1 + tan2 θ = sec2 θ = (tan^4θ + 1)/(𝒕𝒂𝒏𝜽 ×〖𝐬𝐞𝐜〗^𝟐𝜽 ) = (tan^4θ + 1)/(sinθ/cosθ × 1/cos^2θ ) = (tan^4θ + 1)/(sinθ/cos^3θ ) = ((sin𝜃/cos𝜃 )^4 + 1)/(sinθ/cos^3θ ) = (sin^4𝜃/cos^4𝜃 + 1)/(sinθ/cos^3θ ) = ((sin^4𝜃 + cos^4𝜃)/cos^4𝜃 )/(sinθ/cos^3θ ) = (sin^4𝜃 + cos^4𝜃)/cos^4𝜃 ×cos^3θ/sinθ = (〖𝒔𝒊𝒏〗^𝟒𝜽 + 〖𝒄𝒐𝒔〗^𝟒𝜽)/(𝑐𝑜𝑠𝜃 sinθ ) = ((〖𝐬𝐢𝐧〗^𝟐𝜽 + 〖𝐜𝐨𝐬〗^𝟐𝜽 )^𝟐− 𝟐 〖𝐬𝐢𝐧〗^𝟐𝜽 〖𝐜𝐨𝐬〗^𝟐〖𝜽 〗)/(𝑐𝑜𝑠𝜃 sinθ ) Putting 〖𝐬𝐢𝐧〗^𝟐𝜽 + 〖𝐜𝐨𝐬〗^𝟐𝜽 = 1 = (𝟏^𝟐− 2 sin^2𝜃 cos^2〖𝜃 〗)/(𝑐𝑜𝑠𝜃 sinθ ) = (1 − 2 sin^2𝜃 cos^2〖𝜃 〗)/(𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 ) = 𝟏/(𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽 )−(𝟐 〖𝒔𝒊𝒏〗^𝟐𝜽 〖𝒄𝒐𝒔〗^𝟐〖𝜽 〗)/(𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽 ) = 1/(𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 )−2 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠〖𝜃 〗 = 𝐬𝐞𝐜 𝜽 𝐜𝐨𝐬𝐞𝐜 𝜽−𝟐 𝒔𝒊𝒏𝜽 𝒄𝒐𝒔〖𝜽 〗 = RHS Hence proved