Advertisements
Advertisements
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]`
Advertisements
उत्तर
sec2θ = 1 + tan2θ ......[Fundamental tri. identity]
∴ sec2θ = 1 + `(7/24)^2`
∴ sec2θ = 1 + `49/576`
∴ sec2θ =`(576 + 49)/576`
∴ sec2θ = `625/576`
∴ sec θ = `25/24`
∴ cos θ = `24/25` .......`[cos theta = 1/sectheta]`
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
cot2 A – cos2 A = cos2 A . cot2 A
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
Show that : tan 10° tan 15° tan 75° tan 80° = 1
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
If 5 `tan theta = 4,"write the value of" ((cos theta - sintheta))/(( cos theta + sin theta))`
If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
Prove that cot2θ – tan2θ = cosec2θ – sec2θ
