Advertisements
Advertisements
Question
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]`
Advertisements
Solution
\[\text{sec}^2θ = 1 + \boxed{\text{tan}^2θ}\] ...[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]`
RELATED QUESTIONS
Prove the following trigonometric identities.
`tan theta + 1/tan theta` = sec θ.cosec θ
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
if `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`Prove that `(m + n)^(2/3) + (m - n)^(2/3)`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
If `sec theta = x ,"write the value of tan" theta`.
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
If tanθ `= 3/4` then find the value of secθ.
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Find A if tan 2A = cot (A-24°).
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
