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

Choose the correct alternative: cot θ . tan θ = ? - Geometry Mathematics 2

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प्रश्न

Choose the correct alternative:

cot θ . tan θ = ?

पर्याय

  • 1

  • 0

  • 2

  • `sqrt(2)`

MCQ
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उत्तर

1

cot θ. tan θ = `1/"tan θ"`. tan 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 trigonometric identities:

(i) (1 – sin2θ) sec2θ = 1

(ii) cos2θ (1 + tan2θ) = 1


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


Prove the following trigonometric identities.

`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`


Prove the following trigonometric identities.

if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`


If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z c tan θ, show that `x^2/a^2 + y^2/b^2 - x^2/c^2 = 1`


Prove the following identities:

(cos A + sin A)2 + (cos A – sin A)2 = 2


Prove the following identities:

`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`


If `sec theta + tan theta = x,"  find the value of " sec theta`


Prove the following identity :

`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`


Prove the following identity : 

`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`


If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)


Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.


Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.


Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.


Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.


tan θ cosec2 θ – tan θ is equal to


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 `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1


If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.


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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