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If cos θ = 2425, then sin θ = ? - Geometry Mathematics 2

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प्रश्न

If cos θ = `24/25`, then sin θ = ?

योग

उत्तर

cos θ = `24/25`  ......[Given]

We know that,

sin2θ + cos2θ = 1

∴ `sin^2theta + (24/25)^2` = 1

∴ `sin^2theta + 576/625` = 1

∴ sin2θ = `1 - 576/625`

∴ sin2θ = `(625 - 576)/625`

∴ sin2θ = `49/625`

∴ sin θ = `7/25`    ......[Taking square root of both sides]

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अध्याय 6: Trigonometry - Q.2 (B)

संबंधित प्रश्न

Prove the following trigonometric identities:

`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `


Express the ratios cos A, tan A and sec A in terms of sin A.


Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.


Prove the following trigonometric identities.

`((1 + tan^2 theta)cot theta)/(cosec^2 theta)   = tan theta`


Prove the following identities:

`(cosecA - 1)/(cosecA + 1) = (cosA/(1 + sinA))^2`


Prove the following identities:

`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`


`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`


cosec4θ − cosec2θ = cot4θ + cot2θ


`(cos^3 theta +sin^3 theta)/(cos theta + sin theta) + (cos ^3 theta - sin^3 theta)/(cos theta - sin theta) = 2`


If x= a sec `theta + b tan theta and y = a tan theta + b sec theta ,"prove that" (x^2 - y^2 )=(a^2 -b^2)`


Write the value of `(1 - cos^2 theta ) cosec^2 theta`.


Write the value of `4 tan^2 theta  - 4/ cos^2 theta`


Write True' or False' and justify your answer the following: 

\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.


Prove the following identity : 

`sec^4A - sec^2A = sin^2A/cos^4A`


Prove the following identity : 

`(cosecθ)/(tanθ + cotθ) = cosθ`


If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn 


Evaluate:

sin2 34° + sin56° + 2 tan 18° tan 72° – cot30°


Choose the correct alternative:

1 + cot2θ = ? 


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 the following:

(sin α + cos α)(tan α + cot α) = sec α + cosec α


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