Question

(tan a +1/tan a) = sec a.cosec a prove it


plz guys help me to solve this one


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Answer 1

Answer:

tan a+1/tan a

(tan^2 a+1)/tan a

sec^2 a/tan a

(sec a× sec a)×cos a/sin a

(1/cos a×sec a)×cos a/sin a

cos a will be canceled out

sec a/sin a

sec a ×cosec a

Step-by-step explanation:

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Answer 2

Given :-

$\rm \: tan \: a + \dfrac{1}{tan \: a } = sec \: a \: \: cosec \: a$

To find :-

$\rm L.H.S = R.H.S$

Proof :

$\rm R.H.S = \: sec \: a \: \: cosec \: a$

$\rm L.H.S =tan \: a + \dfrac{1}{tan \: a }$

$\rm \longrightarrow \large tan \: a + \dfrac{1}{tan \: a } \\ \\ \\ \rm \longrightarrow \large\dfrac{ {tan}^{2}a + 1}{tan \: a }\\ \\ \\ \rm \bigg( {tan}^{2}a + 1 = {sec}^{2}a \bigg) \\ \\ \\ \rm \large\longrightarrow \dfrac{ {sec}^{2}a}{tan \: a } \\ \\ \\ \rm{sec }^{2} \: a = \dfrac{1}{ {cos}^{2}a } \: and \: \dfrac{1}{tan \: a} = \dfrac{cos \: a}{sin \: a} \\ \\ \\ \\ \rm \large\longrightarrow \dfrac{1 }{{cos}^{2}a } \times \dfrac{cos \: a}{sin \: a} \\ \\ \\ \\ \rm \large\longrightarrow \dfrac{1 }{{cos} \: {}a } \times \dfrac{1}{sin \: a}\\ \\ \\ \\ \rm \large\longrightarrow sec \: a \times{cosec \: a}\\ \\ \\ \\ \rm \large\longrightarrow sec \: a \: \: {cosec \: a}$

Therefore , LHS = RHS

hence proved