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рдкреНрд░рд╢реНрди
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
рдЙрддреНрддрд░
We have `(cosec theta + cot theta ) = m ............(i)`
Again ,`( cosec theta - cot theta )=n ............(ii)`
ЁЭСБЁЭСЬЁЭСд, ЁЭСЪЁЭСвЁЭСЩЁЭСбЁЭСЦЁЭСЭЁЭСЩЁЭСжЁЭСЦЁЭСЫЁЭСФ (ЁЭСЦ)ЁЭСОЁЭСЫЁЭСС (ЁЭСЦЁЭСЦ), ЁЭСдЁЭСТ ЁЭСФЁЭСТЁЭСб:
`(cosec theta + cot theta ) xx ( cosec theta - cot theta ) = mn`
= >`cosec ^2 theta - cot^2 theta =mn`
= >1= mn `[тИ╡ cosec ^2 theta - cot^2 theta =1]`
∴ mn =1
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Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
If x = a cos θ and y = b cot θ, show that:
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If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
If `sec theta = x ,"write the value of tan" theta`.
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
If sec θ = `25/7`, then find the value of tan θ.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
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 `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
