Advertisements
Advertisements
प्रश्न
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
Advertisements
उत्तर
We have,
\[\sin^2 \theta + \cos^2 \theta = 1\]
\[ \Rightarrow \left( \frac{7}{25} \right)^2 + \cos^2 \theta = 1\]
\[ \Rightarrow \cos^2 \theta = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625}\]
\[ \Rightarrow \cos\theta = \sqrt{\frac{576}{625}} = \frac{24}{25}\]
Now,
\[\tan\theta = \frac{\sin\theta}{\cos\theta}\]
\[ \Rightarrow \tan\theta = \frac{\frac{7}{25}}{\frac{24}{25}}\]
\[ \Rightarrow \tan\theta = \frac{7}{24}\]
Thus, the values of cosθ and tanθ are \[\frac{24}{25}\] and \[\frac{7}{24}\], respectively.
संबंधित प्रश्न
If \[\tan \theta = \frac{3}{4}\], find the values of secθ and cosθ
Prove that:
cos2θ (1 + tan2θ)
Prove that:
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?
Choose the correct alternative answer for the following question.
1 + tan2 \[\theta\] = ?
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following.
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Show that:
`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`
In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.
ΔAMT∼ΔAHE, construct Δ AMT such that MA = 6.3 cm, ∠MAT=120°, AT = 4.9 cm and `"MA"/"HA"=7/5`, then construct ΔAHE.
Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
