Advertisements
Advertisements
प्रश्न
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
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 \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
Prove that:
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.
1 + tan2 \[\theta\] = ?
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Prove the following.
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.
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.
