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प्रश्न
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square ...`[cos theta = 1/sectheta]`
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उत्तर
\[\text{sec}^2θ = 1 + \boxed{\text{tan}^2θ}\] ...[Fundamental tri. identity]
∴ \[\text{sec}^2θ = 1 + \boxed{\frac{7}{24}}^2\]
∴ \[\text{sec}^2θ = 1 + \frac{\boxed{49}}{576}\]
∴ sec2θ =`(576 + 49)/576`
∴ \[\text{sec}^2θ = \frac{\boxed{625}}{576}\]
∴ \[\text{sec} \phantom{.}θ = \boxed{\frac{25}{24}}\]
∴ \[\text{cos} \phantom{.}θ = \boxed{\frac{24}{25}}\] ...`[cos theta = 1/sectheta]`
संबंधित प्रश्न
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Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove that:
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Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
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If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
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If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
