Math, asked by rsthakur33202, 1 year ago

1-sin2A/cos2A=1-tanA/1+tanA

Answers

Answered by jomolmary8
39
1-sin2A/cos2A = 1-2sinA.cosA/cos^2A-sin^2A

=cos^2A +sin^2A-2sinAcosA/
cos^2A-sin^2A

= cos^2A-2sonAcosA+sin^2A/
cos^2A-sin^2A

= (cosA +sinA)^2/ (cosA - sinA)
(cosA+sinA)

= (cosA +sinA) (cosA+sinA)/
(cosA-sinA)(cosA+sinA)

= [(cosA+sinA)/(cosA-sinA)]÷cosA

=1+(sinA/cosA) /1-(sinA/cosA)

=1+tanA/1-tanA.
hence it is proved.
Answered by KajalBarad
0

Answer:

1+tanA/1-tanA

Step-by-step explanation:

Given:

Trigonometric equation 1-sin2A/cos2A=1-tanA/1+tanA

To find:

The value of the trigonometric equation 1-sin2A/cos2A=1-tanA/1+tanA

Solution:

Sine Angle: Sine is a trigonometric function of an angle defined as the length of the perpendicular (opposite) side divided by the length of the hypotenuse.

Cosine Angle: The trigonometric function equal to the side adjacent to an acute angle (in a right-angled triangle) divided by the hypotenuse.

1-sin2A/cos2A = 1-2sinA.cosA/cos^2A-sin^2A

=cos^2A +sin^2A-2sinAcosA/cos^2A-sin^2A

= cos^2A-2sonAcosA+sin^2A/cos^2A-sin^2A

= (cosA +sinA)^2/ (cosA - sinA)(cosA+sinA)

= (cosA +sinA) (cosA+sinA)/(cosA-sinA)(cosA+sinA)

= [(cosA+sinA)/(cosA-sinA)]÷cosA

=1+(sinA/cosA) /1-(sinA/cosA)

=1+tanA/1-tanA.

Hence, the value of 1-sin2A/cos2A=1-tanA/1+tanA = 1+tanA/1-tanA.

#SPJ2

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