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Question
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
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Solution
We need to prove `(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
Here, we will first solve the L.H.S.
Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta` we get
`(sec A - tan A)^2 = (1/cos A - sin A/cos A)^2`
`= ((1 -sin A)/cos A)^2`
`= (1 - sin A)^2/(cos A)^2`
Further using the property `sin^2 theta + cos^2 theta = 1` we get
`((1 - sin A)^2/(cos A)) = (1 - sin A)^2/(1 - sin^2 A)`
`= (1 - sin A)^2/((1 - sin A)(1 + sin A))` (using `a^2 - b^2 = (a + b)(a - b))`
`= (1 - sin A)/(1 + sin A)`
henc e proved
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Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
