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प्रश्न
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`
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उत्तर
sec2θ = 1 + tan2θ ......[Fundamental trigonometric identity]
∴ sec2θ = 1 + `(9/40)^2`
∴ sec2θ = 1 + `81/1600`
∴ sec2θ = `1681/1600`
∴ sec θ = `41/40`
संबंधित प्रश्न
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
`sin^2 theta + 1/((1+tan^2 theta))=1`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
Write the value of cosec2 (90° − θ) − tan2 θ.
Find A if tan 2A = cot (A-24°).
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
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)
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
