Advertisements
Advertisements
प्रश्न
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Advertisements
उत्तर
L.H.S. = sec6x – tan6x
L.H.S. = (sec2x)3 – tan6x
L.H.S. = (1 + tan2x)3 – tan6x ...[∵ 1 + tan2θ = sec2θ]
L.H.S. = 1 + 3tan2x + 3(tan2x)2 + (tan2x)3 – tan6x ...[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
L.H.S. = 1 + 3tan2x(1 + tan2x) + tan6x – tan6x
L.H.S. = 1 + 3tan2x.sec2x ...[∵ 1 + tan2θ = sec2θ]
R.H.S. = 1 + 3sec2x.tan2x
L.H.S. = R.H.S.
∴ sec6x – tan6x = 1 + 3sec2x × tan2x
Hence proved.
APPEARS IN
संबंधित प्रश्न
If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.
Prove that:
cos2θ (1 + tan2θ)
Prove that:
Prove that:
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
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.
cosec 45° =?
Choose the correct alternative answer for the following question.
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Choose the correct alternative:
sinθ × cosecθ =?
If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.
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
