Question

16. यदि 9 cos A+ 40 sin A=41 हो, तो cos A और cosec A का मान ज्ञात करें ​

Answer 1

Answer:

$\boxed{\bold{\pink{cos \alpha = \dfrac{9}{41} \: and \: cosec \alpha = \dfrac{41}{40} }}}$

Step-by-step explanation:

$\green{\bold{Given}}\longrightarrow \\ 9 \: cos \alpha + 40 \: sin \alpha = 41$

$\red{\bold{To \: find}}\longrightarrow \\ value \: of \: cos \alpha \: and \: cosec \alpha$

$\blue{\bold{Concept \: used }}\longrightarrow\\ 1) {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ 2) {sin}^{2} \alpha = 1 - {cos}^{2} \alpha \\ 3) {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ 4) \: cosec \alpha = \dfrac{1}{sin \alpha }$

$\pink{\bold{Solution}}\longrightarrow \\ 9cos \alpha + 40sin \alpha = 41$

$= > 40sin \alpha = 41 - 9cos \alpha$

$squaring \: both \: sides \: we \: get$

$= > {(40sin \alpha )}^{2} = {(41 - 9cos \alpha )}^{2}$

$= > 1600 {sin}^{2} \alpha = {(41)}^{2} + {(9cos \alpha) }^{2} - 2 \: (41) \: (9cos \alpha ) \\ = > 1600(1 - {sin}^{2} \alpha ) = 1681 + 81 {cos}^{2} \alpha - 738cos \alpha$

$= > 1600 - 1600 {cos}^{2} \alpha - 81 {cos}^{2} \alpha + 738cos \alpha - 1681 = 0 \\ = > - 1681 {cos}^{2} \alpha + 738cos \alpha + 81 = 0$

$= > 1681 {cos}^{2} \alpha - 738cos \alpha + 81 = 0 \\ = > {(41cos \alpha )}^{2} - 2 \: (41cos \alpha ) \: ( \: 9 \: ) + {(9)}^{2} = 0 \\ = > {(41cos \alpha - 9)}^{2} = 0 \\ taking \: square \: root \: of \: both \: sides \\ = > 41cos \alpha - 9 = 0 \\ = >41 cos \alpha = 9 \\ = >\pink{ cos \alpha = \dfrac{9}{41}} \\ now \\ {sin}^{2} \alpha = 1 - {cos}^{2} \alpha \\ = > sin \alpha = \sqrt{1 - {cos}^{2} \alpha } \\ = > sin \alpha = \sqrt{1 - {( \dfrac{9}{41}) }^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 - \dfrac{81}{1681} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ \dfrac{1681 - 81}{81} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ \dfrac{1600}{1681} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{40}{41}$

$now$

$cosec \alpha = \dfrac{1}{sin \alpha } \\ \\= > cosec \alpha = \dfrac{1}{ \dfrac{ 40}{41} } \\ \\ = >\pink{ cosec \alpha = \dfrac{41}{40}}$

Answer 2

given

9cosA + 40sinA=41

40sinA = 41 - 9cosA

squaring both sides we get

(40sinA)² =(41-9cosA)²

1600sin²A = 1681 + 81cos²A - 2×41×9cosA

1600 - 1600cos²A =1681 + 81cos²A - 738cosA

1681cos²A - 738cosA +81 = 0

(41cosA)² - 2×41cosA×9 + (9)² = 0

(41cosA - 9)² = 0

41cosA - 9 = 0

41cosA = 9

cosA =9/41

we know that

sinA=√1 - cos²A

sinA=√1 - (81/1681)

sinA=√1600/1681

sinA=40/41