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प्रश्न
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove the following:
`(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
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उत्तर
LHS = `(sec θ - tan θ)/(sec θ + tan θ )`
= `(sec θ - tan θ)/(sec θ + tan θ ) xx (sec θ - tan θ)/(sec θ - tan θ )`
= `(sec θ - tan θ)^2/(sec^2θ - tan^2θ )`
= `(sec^2θ + tan^2θ - 2sec θ.tan θ )/1`
= 1 + 2 tan2θ − 2 sec θ. tan θ
= R.H.S.
Hence proved.
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संबंधित प्रश्न
Prove that: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
Find the value of sin ` 48° sec 42° + cos 48° cosec 42°`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
