Question

plz solve these questions .. that r in the pic .. and. solve these on notebook .... n then send me pic .. plz it' s urgent .. frnds

Answer 1

Heyy !
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Q 1 ) $\frac{tan45}{cosec} + \frac{sec60}{cot45} - \frac{3sin60}{2cos0}$

= $\frac{1}{2} + \frac{2}{1} - \frac{3 X \frac{ \sqrt{3} }{2} }{2X1}$

= $\frac{5}{2} - \frac{3 \sqrt{3} }{4}$

= $\frac{10 - 3 \sqrt{3} }{4}$
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Q 2 find $\frac{cos80}{sin10} +cosec31.cos59$

= $\frac{cos80}{sin(90-80)} + cosec(90-59) . cos 59$

= $\frac{cos80}{cos80} + sec59.cos59$

= 1+ 1 

= 2 
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Q3 if √3 tanA =3sinA , find  sin²A-cos²A

= $\sqrt{3} \frac{sin A}{cosA} = 3 sinA$

= $\frac{ \sqrt{3}sinA }{3sinA} =cosA$

= $cos A = \frac{1}{ \sqrt{3} }$

SBS

= $sin^2A = \frac{2}{3}$

= $sin^2 = 1- \frac{1}{3}$

= $cos^2A = \frac{1}{3}$

Now,

=> sin²A-cos²A

= $\frac{2}{3} - \frac{1}{3}$

= $\frac{1}{3}$
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Q4  if cosec A = 12/13 find $\frac{2sinA-cosA}{4sinA - 9cosA}$
Here , 
cosec A = $\frac{13}{12}$

=> $sinA = \frac{1}{cosecA} = \sqrt{1-sin^2A} = \sqrt{1- \frac{12^2}{13^2} } = \sqrt{1- \frac{144}{169} } = \frac{5}{13}$


= $\frac{2sinA-cosA}{4sinA - 9cosA}$

= $\frac{2X \frac{12}{13}-3X \frac{5}{13} }{4X \frac{12}{13} -9X \frac{5}{13} }$

= $\frac{ \frac{24-15}{13} }{ \frac{48-45}{13} }$

= $\frac{24 - 15}{48-45}$

= $\frac{9}{3}$

= 3 
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Q-5 12 sec A = 13 . Find all Trigonometric ratios
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here sec A = 13/12
We know that,
sec A = 1/cos A =  cos A = 12/13

Also cos A = b/h 

Here b = 12 ,h = 13


h²  = b² + p² 

p² = h² - b²

p² = 13² -12²

p² = 25 

p = 5 
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Now , 
Sin A = p/h = 5/13 

cosec A = 1/sin A  = 13/5

cos A = 12/ 13 , sec A = 13/12

tan A = sin A / cos A = 5/12 

cot A = 1/ tan A  = 12 /5
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Q-6 In ΔOPQ, Right angled at P. OP = 7 cm , OQ -PQ = 1 . Find sin∅ and cos ∅
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Answer - In ΔOPQ 
H = OQ 
B = PO
P = PQ
Also OQ  - PQ = 1 (given)______(1)
we have OP = 7 cm 
Also         H² = B² + P²
            OQ²  = OP² + PQ²
            OQ² -PQ² =7²
  ( OQ+PQ) ( OQ-PQ ) = 49 
    1(OQ+PQ)               = 49__________(2)
By 1 and 2 
we get 
2 OQ = 50 
OQ = 25 cm 
Also 
25- PQ = 1 
PQ = 24 cm
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Sin ∅ =  P/H= PO / OQ =  7 / 25 cm
cos ∅ =  B/H = QP / OQ = 24/25
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SahilLincolN