Advertisements
Advertisements
Question
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
Advertisements
Solution
LHS = `1/"sec θ − tan θ"`
LHS = `1/"sec θ − tan θ" × "sec θ + tan θ"/"sec θ + tan θ"`
LHS = `"sec θ + tan θ"/((sec θ − tan θ)(sec θ + tan θ))`
LHS = `(sec θ + tan θ)/(sec^2θ − tan^2θ) ...[(a + b)(a - b) = a^2 - b^2]`
LHS = `(sec θ + tan θ)/1 ...{(1 + tan^2θ = sec^2θ),(∴ sec^2θ − tan^2θ = 1):}`
LHS = sec θ + tan θ
RHS = sec θ + tan θ
LHS = RHS
Hence proved.
APPEARS IN
RELATED QUESTIONS
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
If \[\tan \theta = \frac{3}{4}\], find the values of secθ and cosθ
If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.
Prove that:
cos2θ (1 + tan2θ)
Prove that:
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:
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ
Prove the following.
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Choose the correct alternative:
sinθ × cosecθ =?
Show that:
`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`
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
