Question

please answer this...from maths 10th​

Answer 1

Step-by-step explanation:

i hope it may help you ...

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{cos \theta \: = \: \dfrac{2ab}{ {a}^{2} + {b}^{2} } } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{sin \: \theta}  \\ &\sf{tan \: \theta}  \\ &\sf{sin \: \theta \: . \: tan \: \theta} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\large \boxed{\tt :  ⟼\begin{gathered}\bf\red{ {sin}^{2} \: \theta \: + {cos}^{2} \: \theta \: = \: 1 }\end{gathered}}$

$\large \boxed {\red{\tt :  ⟼ \: tan \: \theta \: = \: \dfrac{sin \: \theta}{cos \: \theta} }}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt :  ⟼ \: {cos} \: \theta \: = \: \dfrac{2ab}{ {a}^{2} + {b}^{2} }$

$\tt :  ⟼ \: {sin}^{2} \theta + {cos}^{2} \theta = 1$

$\tt :  ⟼ \: {sin}^{2} \theta \: = 1 - {cos}^{2} \theta$

$\tt :  ⟼ \: = 1 - { \bigg(\dfrac{2ab}{ {a}^{2} + {b}^{2} } \bigg)}^{2}$

$\tt :  ⟼ \: = 1 - \bigg(\dfrac{4 {a}^{2} {b}^{2} }{ {( {a}^{2} + {b}^{2} )}^{2} } \bigg)$

$\tt :  ⟼ \: = \dfrac{ {( {a}^{2} + {b}^{2} ) }^{2} - 4 {a}^{2} {b}^{2} }{ {( {a}^{2} + {b}^{2} ) }^{2} }$

$\tt :  ⟼ \: = \dfrac{ {a}^{4} + {b}^{4} + 2 {a}^{2} {b}^{2} - 4 {a}^{2} {b}^{2} }{ {( {a}^{2} + {b}^{2}) }^{2} }$

$\tt :  ⟼ \: = \dfrac{ {a}^{4} + {b}^{4} - 2 {a}^{2} {b}^{2} }{ {( {a}^{2} + {b}^{2}) }^{2} }$

$\tt :  ⟼ \: = \dfrac{ {( {a}^{2} - {b}^{2}) }^{2} }{ {( {a}^{2} + {b}^{2} ) }^{2} }$

$\tt :  \implies \: sin \: \theta \: = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

$\large \boxed {\red{\tt :  ⟼ \: Now, \: tan \: \theta = \dfrac{sin \: \theta}{cos \: \theta} }}$

$\tt :  ⟼ \: \therefore \: tan \: \theta = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } \div \dfrac{2ab}{ {a}^{2} + {b}^{2} }$

$\bf\implies \:tan \: \theta \: = \dfrac{ {a}^{2} - {b}^{2} }{2ab}$

${\red{\tt :  ⟼Hence \: sin \: \theta \: \times \: tan \: \theta}}$

$\tt :  ⟼ \: = \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } \times \dfrac{ {a}^{2} - {b}^{2} }{2 ab }$

$\tt :  ⟼ = \dfrac{ {( {a}^{2} - {b}^{2}) }^{2} }{2ab( {a}^{2} + {b}^{2}) }$

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$\large \red{\bf \: ⟼ Explore \: more } ✍$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1