Question

sina/sin(90-a)+cosa/cos(90-a) = sec× cosec = 2cosec2a​

Answer 1

I hope this answer helps you...!!!

Answer 2

$\underline{\underline{\mathfrak{\green{Correction:-}}}}$

Prove:

$\mathsf{\dfrac{sinA}{sin(90-A)}+ \dfrac{cosA}{cos(90-A)} =secA × CosecA </strong><strong>}</strong><strong>$

$\underline{\underline{\mathfrak{\green{Solution:-}}}}$

$\underline{\underline{\pink{To \: prove}}}$

$\mathsf{\dfrac{sinA}{sin(90-A)}+ \dfrac{cosA}{cos(90-A)} =secA × CosecA }$

$\:$

$\underline{\underline{\pink{Proof}}}$

Let us consider LHS,

LHS:

$\mathsf{= \dfrac{sinA}{sin(90-A)}+ \dfrac{cosA}{cos(90-A)} }$

As we know,

$\boxed{\red{sin(90-A) = cosA}}[tex]</p><p></p><p>[tex] \:$

$\boxed{\red{cos(90-A) = sinA}}[tex]</p><p>[tex]$

By sub. the values

$\mathsf{</strong><strong>=</strong><strong> </strong><strong>\dfrac{sinA}{</strong><strong>cosA</strong><strong>}+ \dfrac{cosA}{</strong><strong>sinA</strong><strong>}}$

$\mathsf{= \dfrac{{sin}^{2}A+{cos}^{2}A}{sinA \times cosA}}$

$\:$

$\boxed{\red{{sin}^{2}A+{cos}^{2}A= 1}}[tex]</p><p>[tex]$

$\mathsf{= \dfrac{1}{sinA \times cosA}}$

$\mathsf{= \dfrac{1}{sinA} \times \dfrac{1}{cosA}}$

$\mathsf{= cosecA \times sec A}$

RHS = $\mathsf{= cosecA \times sec A}$

Here,

$\boxed{\</strong><strong>green</strong><strong>{</strong><strong>L</strong><strong>H</strong><strong>S</strong><strong> </strong><strong>=</strong><strong> </strong><strong>RHS</strong><strong>}}[tex]</strong></p><p></p><p></p><p>[tex]$

HENCE PROVED!

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