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Question
Choose the correct alternative answer for the following question.
1 + tan2 \[\theta\] = ?
Options
cot2θ
cosec2θ
sec2θ
tan2θ
Solution
\[1 + \tan^2 \theta = \sec^2 \theta\]
Hence, the correct answer is sec2θ .
RELATED QUESTIONS
If \[\tan \theta = \frac{3}{4}\], find the values of secθ and cosθ
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:
cos2θ (1 + tan2θ)
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
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.
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ
Prove the following.
Prove the following.
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.
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
