मराठी

(secA + tanA) (1 − sinA) = ______. - Mathematics

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प्रश्न

(secA + tanA) (1 − sinA) = ______.

पर्याय

  • sec A

  • sin A

  • cosec A

  • cos A

MCQ
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उत्तर

(secA + tanA) (1 − sinA) = cos A.

Explanation:

(secA + tanA) (1 − sinA)

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

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

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

= `(cos^2A)/cos A`

= cosA

Hence, alternative cosA is correct.

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पाठ 8: Introduction to Trigonometry - Exercise 8.4 [पृष्ठ १९३]

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एनसीईआरटी Mathematics [English] Class 10
पाठ 8 Introduction to Trigonometry
Exercise 8.4 | Q 4.3 | पृष्ठ १९३

संबंधित प्रश्‍न

Prove that:

`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`


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

     


If x =  a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`


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


Prove that:

`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.


Prove the following identity :

`(cotA + tanB)/(cotB + tanA) = cotAtanB`


Prove the following identity :

`1/(tanA + cotA) = sinAcosA`


prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`


Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.


Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`


Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ)  +  cos2 θ.


If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ


Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ


Choose the correct alternative:

sec2θ – tan2θ =?


Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec"  theta)` = sec θ


The value of tan A + sin A = M and tan A - sin A = N.

The value of `("M"^2 - "N"^2) /("MN")^0.5`


Complete the following activity to prove:

cotθ + tanθ = cosecθ × secθ

Activity: L.H.S. = cotθ + tanθ

= `cosθ/sinθ + square/cosθ`

= `(square + sin^2theta)/(sinθ xx cosθ)`

= `1/(sinθ xx  cosθ)` ....... ∵ `square`

= `1/sinθ xx 1/cosθ`

= `square xx secθ`

∴ L.H.S. = R.H.S.


Prove the following that:

`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ


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` 


Statement 1: sin2θ + cos2θ = 1

Statement 2: cosec2θ + cot2θ = 1

Which of the following is valid?


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