Ex 8.3
Ex 8.3, 2 Important You are here
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
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, 2 Write all the other trigonometric ratios of โ A in terms of sec A. cos A We know that cos A = ๐/๐๐๐โก๐จ tan A We know that 1 + tan2 A = sec2 A tan2 A = sec2 A โ 1 tan A = ยฑ โ(๐ ๐๐2 ๐ด โ1) Here, A is acute angle (i.e. less than 90ยฐ) & tan A is positive when A is acute โด tan A = โ(๐๐๐๐ ๐จ โ๐) & tan A is positive when A is acute โด tan A = โ(๐๐๐๐ ๐จ โ๐) cot A cot A = 1/(๐ก๐๐ ๐ด) Putting value of tan A found above = ๐/(โ(ใ๐ฌ๐๐ใ^๐โกใ ๐จ โ๐ใ ) ) cosec A We know that 1 + cot2 A = cosec2 A cosec2 A = 1 + cot2 A Putting value of cot A = ๐/(โ(ใ๐๐๐ใ^๐โกใ ๐จ โ๐ใ ) ) cosec2 A = 1 + (1/(โ(sec^2โกใ๐ด โ 1ใ ) ))^2 cosec2 A = 1 + (1/(๐ ๐๐2 ๐ด โ 1)) cosec2 A = ((1 ร (๐ ๐๐2 ๐ด โ 1) + 1)/(๐ ๐๐2 ๐ด โ1)) cosec2 A = (( ๐ ๐๐2 ๐ด โ 1 + 1)/(๐ ๐๐2 ๐ด โ 1)) cosec2 A = ( ๐ ๐๐2 ๐ด )/(๐ ๐๐2 ๐ด โ 1) cosec A = ยฑ โ(( ๐ ๐๐2 ๐ด )/(๐ ๐๐2 ๐ด โ 1)) cosec A = ยฑ secโก๐ด/โ(๐ ๐๐2 ๐ด โ 1 ) Here, A is acute angle(i.e. less than 90ยฐ) & cosec A is positive when A is acute โด cosec A = secโก๐ด/โ(๐ ๐๐2 ๐ด โ 1 ) sin A We know that sin A = 1/(๐๐๐ ๐๐ ๐ด) Putting value of cosec A from above = 1/((secโก๐ด/โ(sec^2โกใ๐ด โ1ใ )) ) = โ(๐๐๐๐ ๐จ โ ๐)/๐๐๐โก๐จ