Ex 8.3, 4 (iii) - Chapter 8 Class 10 Introduction to Trignometry

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Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii)tan⁡θ/(〖1 − cot〗⁡θ " " )+cot⁡θ/(1 − tan⁡θ ) =1+ sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] Taking L.H.S tan⁡θ/(〖1 − cot〗⁡θ " " )+cot⁡θ/(1 − tan⁡θ ) Converting everything into sin θ and cos θ = (sin⁡θ/cos⁡θ )/(1 − (cos⁡θ/sin⁡θ ) )+((cos⁡θ/sin⁡θ ))/(1 − (sin⁡θ/cos⁡θ ) ) = (sin⁡θ/cos⁡θ )/(((sin⁡θ − cos⁡θ)/sin⁡θ ) )+((cos⁡θ/sin⁡θ ))/( ((cos⁡θ − sin⁡θ)/cos⁡θ ) ) = sin⁡θ/cos⁡θ × sin⁡θ/(sin⁡θ −〖 cos〗⁡θ ) +cos⁡θ/sin⁡θ ×cos⁡θ/(cos⁡θ −〖 sin〗⁡θ ) = sin2⁡θ/(〖cos⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) + cos2⁡θ/(sin⁡〖θ 〗⁡〖(𝐜𝐨𝐬⁡𝜽 〗 −〖 𝒔𝒊𝒏〗⁡〖𝜽 )〗 ) = sin2⁡θ/(〖cos⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) + cos2⁡θ/(〖sin⁡〖θ ×〗 − ( 𝐬𝐢𝐧〗⁡𝜽 −〖 𝒄𝒐𝒔〗⁡〖𝜽 )〗 ) = sin2⁡θ/(〖cos⁡θ (sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) − cos2⁡θ/(〖sin⁡θ ( sin〗⁡θ − cos⁡〖θ )〗 ) = (𝑠𝑖𝑛2⁡𝜃 × (𝑠𝑖𝑛⁡𝜃 ) −〖 𝑐𝑜𝑠2〗⁡𝜃 × (𝑐𝑜𝑠⁡𝜃 ))/(〖𝑐𝑜𝑠⁡𝜃 ( 𝑠𝑖𝑛〗⁡𝜃 −〖 𝑐𝑜𝑠〗⁡〖𝜃 )𝑠𝑖𝑛⁡𝜃 〗 ) = (𝐬𝐢𝐧𝟑⁡𝜽 − 𝐜𝐨𝐬𝟑⁡𝜽 )/(〖cos⁡θ sin⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ)〗 ) Since a3 – b3 = (a – b) (a2 + b2 + ab) Putting a = sin θ , b = cos θ = (〖(sin〗⁡θ − cos⁡θ)〖(sin2〗⁡θ+ cos2⁡〖θ +cos⁡θ sin⁡θ 〗))/(〖cos⁡θ sin⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ ) 〗 ) = (〖(sin2〗⁡θ + cos2⁡〖θ + cos⁡θ sin⁡θ 〗))/(cos⁡θ sin⁡θ ) = (〖(𝐬𝐢𝐧𝟐〗⁡𝜽 + 𝐜𝐨𝐬𝟐⁡〖𝜽 ) +〖 cos〗⁡θ sin⁡θ 〗)/(cos⁡θ sin⁡θ ) As cos2 A + sin2 A = 1 = (𝟏 + cos⁡θ sin⁡θ)/(cos⁡θ sin⁡θ ) = (1 )/(cos⁡θ sin⁡θ ) + (cos⁡θ sin⁡θ)/(cos⁡θ sin⁡θ ) = (𝟏 )/𝒄𝒐𝒔⁡𝜽 × (𝟏 )/𝒔𝒊𝒏⁡𝜽 + 1 = sec θ × cosec θ + 1 = 1 + sec θ cosec θ = R.H.S ∴ L.H.S = R.H.S Hence proved