Advertisements
Advertisements
प्रश्न
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
Advertisements
उत्तर
`(sin 50°)/(cos 40 °)+ ( cosec 40° )/( sec 50°) - 4 cos 50° cosec 40°`
`=(cos (90°- 50°))/(cos 40°) + (sec (90°- 40°))/(sec 50°)- 4 sin (90°-50°) cosec 40°`
`=(cos 40° )/( cos 40 °) + ( sec50°)/( sec 50°) - 4 sin 40 ° xx 1/ ( sin 40 °)`
= 1 + 1 - 4
= - 2
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`
`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2 = 2`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
`(1-cos^2theta) sec^2 theta = tan^2 theta`
`1+(tan^2 theta)/((1+ sec theta))= sec theta`
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = cosec θ - cot θ`.
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
The value of sin2θ + `1/(1 + tan^2 theta)` is equal to
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that sec2θ – cos2θ = tan2θ + sin2θ
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
