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Maths-

General

Easy

Question

cosec $straight A equals 4 straight p plus fraction numerator 1 over denominator 16 straight p end fraction not stretchy rightwards double arrow$ cosec A +cot A=

8p
$\frac{1}{8 p}$
$- 8 p (o r) \frac{1}{8 p}$
$8 p (or) \frac{1}{8 p}$

The correct answer is: $8 straight p space left parenthesis or right parenthesis space fraction numerator 1 over denominator 8 straight p end fraction$

$S l o p e blank o f blank 3 x plus 4 y plus 5 equals 0 blank i s blank fraction numerator negative 3 over denominator 4 end fraction blank semicolon left parenthesis 1 comma negative 2 right parenthesis$
$e r blank s l o p e equals fraction numerator 4 over denominator 3 end fraction equals t a n theta$
$open parentheses x comma y close parentheses equals open parentheses x subscript 1 end subscript plus r c o s theta comma y subscript 1 end subscript plus r c o s theta close parentheses$
$equals blank open parentheses 1 plus fraction numerator 3 r over denominator 5 end fraction minus 2 plus fraction numerator 4 r over denominator 5 end fraction close parentheses blank l i e s blank o n blank 2 x plus 3 y minus 14 equals 0$
$equals 2 open parentheses 1 plus fraction numerator 3 r over denominator 5 end fraction close parentheses plus 3 open parentheses negative 2 plus fraction numerator 4 r over denominator 5 end fraction close parentheses minus 14 equals 0$
$r equals 5$
$therefore c e n t r e blank left parenthesis 42 right parenthesis$
$e r blank s l o p e equals negative fraction numerator 5 over denominator 12 end fraction equals t a n theta equals blank theta blank ϵ blank I I Q$
$s i n theta equals fraction numerator 5 over denominator 13 end fraction blank semicolon c o s theta equals fraction numerator negative 12 over denominator 13 end fraction$
$C subscript 1 end subscript C subscript 2 end subscript equals open parentheses x subscript 1 end subscript plus r c o s theta comma y subscript 1 end subscript plus r s i n theta close parentheses$
$equals blank open parentheses 2 plus-or-minus 13. open parentheses fraction numerator negative 12 over denominator 13 end fraction close parentheses comma blank 4 plus 13 open parentheses fraction numerator 5 over denominator 13 end fraction close parentheses close parentheses$
$equals open parentheses negative 109 close parentheses comma open parentheses 14 comma negative 1 close parentheses$

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