Question

secA divided by tanA + cotA is equal to?​

Answer 1

GIVEN :-

• {sec A/(tan A + cot A)}

TO FIND :-

• The value of {sec A/(tan A + cot A)}.

SOLUTION :-

⇒ {sec A/(tan A + cot A)}

As we know that the identity :- tan A = sin A/cos A and also cot A = cos A/sin A.

⇒ [ sec A/{ (sin A/cos A) + (cos A/sin A) } ]

By cross multiplication we get,

⇒ sec A/{ (sin A . sin A + cos A . cos A)/cos A sin A}

⇒ sec A/{ (sin² A + cos² A)/sin A cos A}

As we know that the identity :- sin² A + cos² A = 1

⇒ sec A/(1/cos A sin A)

As we know that the identity :- sec A = 1/cos A.

⇒ (1/cos A) × (cos A sin A/1)

⇒ sin A.

Hence the value of {sec A/(tan A + cot A)} is sin A.

Answer 2

Given:

$\sf \to \dfrac{ \sec A }{\tan A + \cot A}$

Find:

$\sf \to Value \: of \: \dfrac{ \sec A }{\tan A + \cot A}$

Solution:

we, have

$\sf \to \dfrac{ \sec A }{\tan A + \cot A}$

we, know that

$\boxed{ \sf \tan A = \dfrac{ \sin A}{\cos A}} \: \sf and \: \boxed{ \sf \cot A = \dfrac{ \cos A}{\sin A}}$

Put these values in the Question we, get

$\sf \to \dfrac{ \sec A }{\tan A + \cot A}$

$\sf \to \dfrac{ \sec A }{ \dfrac{ \sin A}{\cos A} + \dfrac{ \cos A}{\sin A}}$

$\qquad\qquad\qquad$ Taking L.C.M

$\sf \to \dfrac{ \sec A }{ \dfrac{ {\sin}^{2} A + {\cos}^{2} A }{\cos A \sin A } }$

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Now, we know that

$\boxed{\sf {\sin}^{2} A + {\cos}^{2} A = 1}$

Using this value we, got

$\sf \to \dfrac{ \sec A }{ \dfrac{ {\sin}^{2} A + {\cos}^{2} A }{\cos A \sin A } }$

$\sf \to \dfrac{ \sec A }{ \dfrac{1}{\cos A \sin A } }$

$\sf \to \sec A \times \dfrac{\cos A \sin A}{1}$

$\sf \to \sec A \times \cos A \sin A$

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By, using

$\boxed{\sf \sec A = \dfrac{1}{\cos A} }$

So, putting this value

$\sf \to \sec A \times \cos A \sin A$

$\sf \to \dfrac{1}{\cos A} \times \cos A \sin A$

$\sf \to \dfrac{1}{ \cancel{\cos A}} \times \cancel{ \cos A} \sin A$

$\sf \to 1 \times \sin A$

$\sf \to \sin A$

_________________

$\therefore\underline{ \sf \dfrac{ \sec A }{\tan A + \cot A} = \sin A }$