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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Choose the correct alternative: cos θ. sec θ = ? - Geometry Mathematics 2

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प्रश्न

Choose the correct alternative:

cos θ. sec θ = ?

पर्याय

  • 1

  • 0

  • `1/2`

  • `sqrt(2)`

MCQ
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उत्तर

1

cos θ. sec θ = cos θ. `1/"cos θ"` = 1.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Trigonometry - Q.1 (A)

संबंधित प्रश्‍न

Prove the following identities:

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


Prove that:

`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`


Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`


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 + cot^2 theta ) sin^2 theta`. 


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

The value of  \[\cos^2 23 - \sin^2 67\]  is positive . 


\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to 


If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ = 


If  cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to  

 


Prove the following identity :

`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`


For ΔABC , prove that : 

`sin((A + B)/2) = cos"C/2`


Choose the correct alternative:

1 + tan2 θ = ?


Evaluate:

`(tan 65^circ)/(cot 25^circ)`


Prove the following identities.

`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ


Prove the following identities.

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


Choose the correct alternative:

1 + cot2θ = ? 


Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B


Prove that

`(cot "A" + "cosec  A" - 1)/(cot"A" - "cosec  A" + 1) = (1 + cos "A")/"sin A"`


Eliminate θ if x = r cosθ and y = r sinθ.


Find the value of sin2θ  + cos2θ

Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB2 + BC2 = `square`   .....(Pythagoras theorem)

Divide both sides by AC2

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 


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