Advertisements
Advertisements
प्रश्न
Prove that:
cos2θ (1 + tan2θ)
उत्तर
L.H.S. = cos2θ (1 + tan2θ)
= cos2θ × sec2θ ...[∵ 1 + tan2 θ = sec2 θ]
= \[\cos^{2}\theta\times\frac{1}{\cos^{2}\theta}\]
= 1
= R.H.S.
∴ cos2θ (1 + tan2θ) = 1
संबंधित प्रश्न
If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.
If tanθ = 1 then, find the value of
`(sinθ + cosθ)/(secθ + cosecθ)`
Prove that:
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
Prove that:
Prove that:
If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]
Prove that:
Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?
Choose the correct alternative answer for the following question.
1 + tan2 \[\theta\] = ?
Choose the correct alternative answer for the following question.
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Choose the correct alternative:
sinθ × cosecθ =?
Show that:
`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`
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