Question

9 identity used in 8.4 class 10
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Answer 1

Answer:

FEW BASICS: (Mentally verify these)

1) sin A x cosec A = 1

2) cos A x sec A = 1

3) cosec A x tan A = sec A

4) sec A x cot A = cosec A

5) sin A x sec A = tan A

6) cos A x cosec A = cot A

Identities:

2) sin2A + cos2A =1 Also: sin2A = 1 - cos2A; cos2A = 1 - sin2A

3) tan2A + 1 = sec2A Dividing all of 2) by cos2A we get this.

4) 1 + cot2A = cosec2A Dividing all of 2) by sin2A we get this

From3) we also get: sec2A - 1 = tan2A and sec2A - tan2A = 1

From4) we also get: cosec2A - 1 = cot2A and cosec2A - cot2A = 1

EXAMPLES:

Example 13: Prove that sec A(1 - sin A)(sec A + tan A) = 1

Solution:

L. H. S. = sec A(1 - sin A)(sec A + tan A)

= (sec A - tan A) (sec A + tan A)

= sec2 A - tan2 A

= 1 = R.H.S

Example 14: Prove

EXERCISE 8.4

or

Note_these simple manipulations

01)

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02)

Multiply Nr and Dr. by (1 + sin A)

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