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

Prove that (sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ).

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प्रश्न

Prove that `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`.

सिद्धांत
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उत्तर

L.H.S. = `(sin θ + tan θ)/(cos θ)`

= `(sin θ)/(cos θ) + (tan θ)/(cos θ)`

= tan θ + tan θ sec θ

= tan θ (1 + sec θ)

= R.H.S.

∴ `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`

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पाठ 6: Trigonometry - Exercise

संबंधित प्रश्‍न

If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`


Prove the following trigonometric identities.

`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`


Prove the following trigonometric identities.

`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`


Prove the following identities:

`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`


Prove the following identities:

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


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


If (tan θ + sin θ) = m and (tan θ – sin θ) = n, prove that (m2 – n2)2 = 16 mn.


Write the value of tan10° tan 20° tan 70° tan 80° .


Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`


Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ


If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ. 


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

The value of the expression \[\sin {80}^° - \cos {80}^°\] 


Prove the following identity : 

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


Prove the following identity : 

`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`


If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that   `x^2 + y^2 + z^2 = r^2`


Evaluate:

sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°


Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ


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


Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`


`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.

Activity:

`5/(sin^2θ) - 5cot^2θ`

= `square (1/(sin^2θ) - cot^2θ)`

= `5(square - cot^2θ)   ...[1/(sin^2θ) = square]`

= 5(1)

= `square`


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