Question

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Answer 1

Answer:

u r answer for the question!!!

Answer 2

Answer:

$\huge\bold{QUESTION}$

✧

$if \: sinθ = \frac{20}{29 } \: then \: find \: cosθ \:$

$\huge\bold{ANSWER}$

Solution:-

We have sin²θ+cos²θ= 1

=>

$= > ( { \dfrac{20}{29}) }^{2} + {cos}^{2} θ = 1 \\ \\ = > \frac{400}{841} + {cos}^{2} θ = 1 \\ \\ = > {cos}^{2} θ = 1 - \frac{400}{841} \\ \\ = > {cos}^{2} θ = \frac{841 - 400}{841} \\ \\ = > {cos}^{2} θ = \frac{441}{841} \\ \\ = > {cos} θ = \sqrt{ \frac{441}{841} } = \frac{21}{29} \\ \\ = > cosθ = \frac{21}{29}$

EXPLANATION

Here The Identity: Sin²θ+Cos²θ=1 is used

We are Given the value of sinθ and asked for cosθ

STEPS

➜Place the value of sinθ from given question in the identity Sin²θ+Cos²θ=1

➜Then find the value of cos²θ by solving LHS and RHS

${cos}^{2} θ= \dfrac{441}{841}$

➜Take square root on both side and hence we would find the value of

$cosθ = \dfrac{21}{29}$

ADDITIONAL INFORMATION

1. Identities

• Sin²θ+Cos²θ=1

• Sec²θ-Tan²θ=1

• Cosec²θ-Cot²θ=1

• Sin(90-θ)=cosθ

• tan(90-θ)=cotθ

• sec(90-θ)=cosecθ

• $sinθ = \frac{1}{cosecθ}$
• $cosθ = \frac{1}{sec θ }$

• $tanθ = \frac{1}{cot}θ$

• sin(α+β)=sin(α)cos(β)+cos(α)sin(β)

• sin(α–β)=sin(α)cos(β)–cos(α)sin(β)

• cos(α+β)=cos(α)cos(β)–sin(α)sin(β)

• cos(α–β)=cos(α)cos(β)+sin(α)sin(β)

• sin(α)cos(β)=½(sin(α+β)+sin(α−β))

• cos(α)cos(β)=½(cos(α−β)+cos(α+β))