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प्रश्न
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
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उत्तर
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
consider `sin42^circ sec48^circ + cos42^circ cosec48^circ`
⇒ `sin42^circ sec(90^circ - 42^circ) + cos42^circ cosec(90^circ - 42^circ)`
⇒ `sin42^circ . cosec42^circ + cos42^circ sec42^circ`
⇒ `sin42^circ . 1/sin42^circ + cos42^circ 1/cos42^circ`
⇒ 1 + 1 = 2
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
Prove the following identities:
`(1 + cos theta - sin^2 theta )/(sin theta (1 + cos theta)) = cot theta`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
