If $space tanA space equals space 5 over 12 comma$ find the value of (sinA + cosA) secA.

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Solution

Let us consider a right triangle right angled at B.
We have,

rightwards double arrow space space tanA space equals space 5 over 12 space equals space AB over BC

If AB = 5K, BC = 12K
Then,

space AC squared equals AB squared plus BC squared space equals space left parenthesis 5 straight K right parenthesis squared plus left parenthesis 12 straight K right parenthesis squared space space space space space space space space space space equals space 25 straight K squared plus 144 straight K squared space equals space 169 straight K squared

So, AC = 13K
Now,

sinA space equals space AB over AC equals fraction numerator 5 straight K over denominator 13 straight K end fraction equals 5 over 13 cosA space equals space BC over AC equals fraction numerator 12 straight K over denominator 13 straight K end fraction space equals 12 over 13

secA space equals space AC over BC equals fraction numerator 13 straight K over denominator 12 straight K end fraction equals 13 over 12

Therefore,

left parenthesis sinA plus cosA right parenthesis space secA space equals space open parentheses 5 over 13 plus 12 over 13 close parentheses cross times 13 over 12 equals space open parentheses fraction numerator 5 plus 12 over denominator 13 end fraction close parentheses cross times 13 over 12 space equals space 17 over 13 cross times 13 over 12 space equals space 17 over 12.

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