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 April 16, 2024 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 = 𝟏/𝒄𝒐𝒕𝑨 cosec A We know that 1 + cot2 A = cosec2 A cosec2 A = 1 + cot2 A cosec A = ± √(1+𝑐𝑜𝑡2 𝐴) Here, A is acute angle (i.e. less than 90°) & cosec A is positive when A is acute ∴ cosec A = √(𝟏+𝒄𝒐𝒕𝟐 𝑨) sin A sin A = 1/(𝑐𝑜𝑠𝑒𝑐 𝐴) Putting value of cosec A found above = 𝟏/(√(𝟏 + 𝒄𝒐𝒕^𝟐 𝑨) ) sec a We know that 1 + tan2 A = sec2 A sec2 A = 1 + tan2 A sec A = ± √((1+𝑡𝑎𝑛2 𝐴)) Here, A is acute angle(i.e. less than 90°) & sec A is positive when A is acute ∴ sec A = √((1+𝑡𝑎𝑛2 𝐴)) Converting tan A to cot A = √((𝟏+𝟏/(𝒄𝒐𝒕𝟐 𝑨))) = √(((𝑐𝑜𝑡2 𝐴 + 1)/(𝑐𝑜𝑡2 𝐴)) ) = √(〖𝒄𝒐𝒕〗^𝟐𝑨 + 𝟏)/𝒄𝒐𝒕𝑨