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

Choose the correct alternative: tan (90 – θ) = ? - Geometry Mathematics 2

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प्रश्न

Choose the correct alternative:

tan (90 – θ) = ?

पर्याय

  • sin θ

  • cos θ

  • cot θ

  • tan θ

MCQ
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उत्तर

cot θ

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

संबंधित प्रश्‍न

 

Evaluate

`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`

 

Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.


Prove the following trigonometric identities.

(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A


Prove the following trigonometric identities.

`(tan A + tan B)/(cot A + cot B) = tan A tan B`


Prove the following identities:

`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`


Prove the following identities:

(1 + tan A + sec A) (1 + cot A – cosec A) = 2


`(1-cos^2theta) sec^2 theta = tan^2 theta`


`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`


`(cos theta  cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`


Write the value of `(1 + tan^2 theta ) cos^2 theta`. 


\[\frac{x^2 - 1}{2x}\] is equal to 


Prove the following identity : 

`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`


Prove the following identity : 

`(1 + tan^2θ)sinθcosθ = tanθ`


Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`


If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1


The value of sin2θ + `1/(1 + tan^2 theta)` is equal to 


If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1


(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.


If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.


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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