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Ex 8.4
Ex 8.4, 2 Important
Ex 8.4, 3 (i) Important
Ex 8.4, 3 (ii)
Ex 8.4, 4 (i) [MCQ]
Ex 8.4, 4 (ii) [MCQ] Important
Ex 8.4, 4 (iii) [MCQ] Important
Ex 8.4, 4 (iv) [MCQ]
Ex 8.4, 5 (i) Important
Ex 8.4, 5 (ii)
Ex 8.4, 5 (iii) Important
Ex 8.4, 5 (iv) Important
Ex 8.4, 5 (v) Important
Ex 8.4, 5 (vi)
Ex 8.4, 5 (vii) Important
Ex 8.4, 5 (viii)
Ex 8.4, 5 (ix) Important
Ex 8.4, 5 (x)
Last updated at Aug. 5, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 8.4, 1 Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. tan A We know that tan A = π/πππβ‘π¨ cosec A We know that 1 + cot2 A = cosec2 A cosec2 A = 1 + cot2 A cosec A = Β± β(1+πππ‘2 π΄) Here, A is acute (i.e. less than 90Β°) & cosec A is positive when A is acute So, cosec A = β(1+πππ‘2 π΄) sin A sin A = π/(πππππ π¨) Putting value of cosec A found above sin A = 1/(β(1 + πππ‘2 π΄) ) Sec A We know that 1 + tan2 A = sec2 A sec2 A = 1 + tan2 A sec A = Β± β((1+π‘ππ2 π΄)) Here, A is acute (i.e. less than 90Β°) & sec A is positive when A is acute sec A = β((1+π‘ππ2 π΄)) sec A = β((1+1/(πππ‘2 π΄))) sec A = β(((πππ‘2 π΄ + 1)/(πππ‘2 π΄)) ) sec A = β(πππ‘2 π΄ + 1)/cotβ‘π΄