Question 23 - If tan θ = 1/√5, then find the value of (cosec² θ

Question 23 If tan⁡𝜃=(1)/(sqrt(5)), then find the value of (cosec^(2)𝜃 − sec^(2)⁡𝜃)/(cosec^(2)𝜃 + sec^(2)⁡𝜃). Given tan𝜃=(1)/(sqrt(5)) Now, 𝐬𝐞𝐜^(𝟐) 𝜽= 𝟏+𝐭𝐚𝐧^(𝟐)𝜽 sec^(2)𝜃=1+((1)/(sqrt(5)))^(2) sec^(2)𝜃=1+(1)/(5) 𝒔𝒆𝒄^(𝟐)𝜽=(𝟔)/(𝟓) We need to find cosec^(2)𝜃 Since tan𝜃=(1)/(sqrt(5)) ∴ 𝐜𝐨𝐭𝜽=sqrt(𝟓) Now, 𝐜𝐨𝒔𝒆𝒄^(𝟐) 𝜽= 𝟏+𝐜𝐨𝐭^(𝟐)𝜽 cosec^(2)𝜃=1+(sqrt(5))^(2) cosec^(2)𝜃=1+5 𝒄𝒐𝒔𝒆𝒄^(𝟐)𝜽=𝟔 Now, (𝒄𝒐𝒔𝒆𝒄^(𝟐)𝜽 − 𝒔𝒆𝒄^(𝟐)⁡𝜽)/(𝒄𝒐𝒔𝒆𝒄^(𝟐)𝜽 + 𝒔𝒆𝒄^(𝟐)⁡𝜽)=(6 − (6)/(5))/(6 + (6)/(5)) =((6 × 5 − 6)/(5))/((6 × 5 + 6)/(5)) =((30 − 6)/(5))/((30 + 6)/(5)) =((30 − 6)/(5))/((30 + 6)/(5)) Converting into sin and cos =((1)/(sin^(2)𝜃) − (1)/(cos^(2)𝜃))/((1)/(sin^(2)𝜃) + (1)/(cos^(2)𝜃)) =((sin^(2)𝜃 − cos^(2)𝜃)/(sin^(2)𝜃cos^(2)𝜃) )/((sin^(2)𝜃 + cos^(2)𝜃)/(sin^(2)𝜃cos^(2)𝜃)) =(𝒔𝒊𝒏^(𝟐)𝜽 − 𝒄𝒐𝒔^(𝟐)𝜽)/(𝒔𝒊𝒏^(𝟐)𝜽 + 𝒄𝒐𝒔^(𝟐)𝜽) Putting 𝑠𝑖𝑛^(2)𝜃 + 𝑐𝑜𝑠^(2)𝜃 = 1 =(sin^(2)𝜃 − cos^(2)𝜃)/(1) =sin^(2)𝜃 − cos^(2)𝜃 Since we know 𝑡𝑎𝑛𝜃, we divide =((6 × 5 − 6)/(5))/((6 × 5 + 6)/(5)) =((30 − 6)/(5))/((30 + 6)/(5)) =(30 − 6)/(30 + 6) =(24)/(36) =(𝟐)/(𝟑)

Question 23 - CBSE Class 10 Sample Paper for 2027 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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