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Ex 8.3

Ex 8.3, 1
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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

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 May 29, 2023 by Teachoo

Ex 8.3, 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΄