Question

सही विकल्प चुनिए और अपने विकल्प की पुष्टि कीजिए:
$left parenthesis 1 plus tanθ plus secθ right parenthesis space left parenthesis 1 plus cotθ minus cosecθ right parenthesis$ बराबर हैं:

0

1

2

-1

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Solution

left parenthesis 1 plus tanθ plus secθ right parenthesis space left parenthesis 1 plus cotθ minus cosecθ right parenthesis

equals space open parentheses 1 plus sinθ over cosθ plus 1 over cosθ close parentheses space open parentheses 1 plus cosθ over sinθ minus 1 over sinθ close parentheses

equals space open parentheses fraction numerator cosθ plus sinθ plus 1 over denominator cosθ end fraction close parentheses space open parentheses fraction numerator sinθ plus cosθ minus 1 over denominator sinθ end fraction close parentheses

equals space fraction numerator open curly brackets left parenthesis sin space straight theta plus space cos space straight theta right parenthesis plus 1 close curly brackets space open curly brackets left parenthesis sinθ plus cosθ right parenthesis minus 1 close curly brackets over denominator cosθ. space sinθ end fraction

space equals space fraction numerator left parenthesis sinθ plus cosθ right parenthesis squared minus left parenthesis 1 right parenthesis squared over denominator cosθ. space sinθ end fraction

equals fraction numerator sin squared straight theta plus cos squared straight theta plus 2 sinθ. cosθ minus 1 over denominator sinθ. cosθ end fraction

equals space fraction numerator 1 plus 2 sinθ. space cosθ minus 1 over denominator sinθ. space cosθ end fraction space equals space fraction numerator 2 space sinθ. space space cosθ over denominator sinθ. space cosθ end fraction equals 2

अत: विकल्प (C) सही हैं।

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