Question

show that coscosA/1-tantanA+sinsinA/1-cotA=sinA+cosA​

Answer 1

Answer:

Proved below:

Step-by-step explanation:

CORRECT QUESTION:

Show that

$\frac{cosA}{1-tanA} +\frac{sinA}{1-cotA}=sinA+cosA$

$=\frac{ cosA } { 1-\frac{sinA}{cosA} } + \frac { sinA } { 1-\frac{cosA}{sinA} }\\ \\ =\frac{ cosA } { \frac{cosA-sinA}{cosA} } + \frac { sinA } { \frac{sinA-cosA}{sinA} }\\ \\\\=\frac{cos^{2}A}{cosA-sinA}+\frac{sin^{2}A}{sinA-cosA}\\\\= \frac{cos^{2}A}{cosA-sinA}+\frac{sin^{2}A}{-[cosA-sinA]}\\\\\=\frac{cos^{2}A}{cosA-sinA}-\frac{sin^{2}A}{cosA-sinA}\\\\=\frac{cos^{2}A-sin^{2}A}{cosA-sinA}\\\\=\frac{(cosA-sinA)(cosA+sinA)}{cosA-sinA}\;\;\;\;$

$=\frac{cos^a}{cosA-sinA}-\frac{cos^a}{cosA-sinA}$

$=\frac{cos^a-sin^{2}A}{cosA-sinA}= \frac{[cosa+sinA][cosA-sinA]}{cosA-sinA}$

{     because   $a^{2}-b^{2}=[a+b][a-b]$       }

$=cosA+sinA=sinA+cosA$

$Learning\;\;Together\;:)$

Answer 2

Answer:

Step-by-step explanation:

LHS

$\frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}$

$\frac{cosA}{1-\frac{sinA}{cosA}}+\frac{sinA}{1-\frac{cosA}{sinA}}$

Using $tanA=\frac{sinA}{cosA}$

$cotA=\frac{cosA}{sinA}$

$\frac{cos^2A}{cosA-sinA}-\frac{sin^2A}{cosA-sinA}$

$\frac{cos^2A-sin^2A}{cosA-sinA}$

$\frac{(cosA-sinA)(cosA+sinA)}{cosA-sinA}$

Using property

$(a^2-b^2)=(a+b)(a-b)$

$cosA+sinA$

LHS=RHS

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https://brainly.in/question/753217 Answered by Qais