Advertisements
Advertisements
प्रश्न
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Advertisements
उत्तर
We have tanA = n tan B
⇒ `cot B = n/tanA ........(i)`
Again , sin A = m sin B
` ⇒ cosec B = m/ sin A ........(ii) `
Squaring (i) and ( ii) and subtracting (ii) from (i) , We get
`⇒ (m^2)/(sin^2 A) - (n^2 )/(tan^2 A) = cosec ^2 B - cot^2 B`
`⇒ (m^2 )/(sin^2 A )-(n^2 cos )/(sin^2 A)m=1`
`⇒m^2 - n^2 cos^2 A = sin^2 A`
`⇒ m^2 - n^2 cos^2 A =1- cos^2 A`
`⇒ n^2 cos^2 A- cos^2 A = m^2 -1`
`⇒cos^2 A (n^2 -1) = (m^2 -1)`
`⇒ cos^2 A = ((m^2 -1))/((n^2 -1))`
∴` cos^2 A = ((m^2 -1))/((n^2 -1))`
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
` tan^2 theta - 1/( cos^2 theta )=-1`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove that:
tan (55° + x) = cot (35° – x)
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
sin(45° + θ) – cos(45° – θ) is equal to ______.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
