Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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

Ex 8.3, 4 (iv) Important

Ex 8.3, 4 (v) Important You are here

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 2024 Exams

Question 1 (ii) Deleted for CBSE Board 2024 Exams

Last updated at Aug. 16, 2023 by Teachoo

Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. "cos A – sin A + 1" /"cos A + sin A – 1" = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A. Solving L.H.S (cos𝐴 − sin𝐴 + 1)/(cos𝐴 + sin𝐴 − 1) Since we need to use cosec and cot identity Dividing both numerator and denominator by sin A = (𝟏/𝒔𝒊𝒏〖 𝑨〗 (cos〖 𝐴 − sin〖𝐴 + 1〗 〗 ))/(𝟏/𝒔𝒊𝒏〖 𝑨〗 (cos〖 𝐴 + sin〖 𝐴 − 1〗 〗 ) ) = (cos〖 𝐴〗/sin〖 𝐴〗 − sin〖 𝐴〗/sin〖 𝐴〗 + 1/sin〖 𝐴〗 )/(cos〖 𝐴〗/sin〖 𝐴〗 + sin〖 𝐴〗/sin〖 𝐴〗 − 1/sin〖 𝐴〗 ) = cot〖 𝐴 − 1 + 𝑐𝑜𝑠𝑒𝑐 𝐴〗/cot〖 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴〗 = ((cot〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − 𝟏〗)/((cot〖 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴) 〗 ) = ((co𝑡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − (𝒄𝒐𝒔𝒆𝒄𝟐 𝑨 − 𝒄𝒐𝒕𝟐 𝑨)〗)/((cot 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴)) = ((co𝑡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − (𝐜𝐨𝒔𝒆𝒄𝑨 − 𝒄𝒐𝒕 𝑨)(𝐜𝐨𝒔𝒆𝒄𝑨 + 𝒄𝒐𝒕 𝑨)〗)/((cot 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴)) = ((co𝑡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) [𝟏 − (𝒄𝒐𝒔𝒆𝒄 𝑨 − 𝒄𝒐𝒕 𝑨 )]〗)/([cot 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴]) = ((co𝑡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) [1 − 𝑐𝑜𝑠𝑒𝑐 𝐴 + 𝑐𝑜𝑡 𝐴]〗)/([cot 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴]) = ((co𝑡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴)[cot 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴]〗)/([cot 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴]) = cot A + cosec A = R.H.S Hence proved