Advertisements
Advertisements
प्रश्न
Prove that
cosec (67° + θ) − sec (23° − θ) = 0
उत्तर
\[\begin{array}{l} \text{L.H.S}=cosec( {67}^\circ + \theta) - \sec( {23}^\circ- \theta) \\ \end{array}\]
\[\begin{array}{l}=cosec{ {90}^\circ - ( {23}^\circ-\theta)}-\sec( {23}^\circ - \theta) \\ \end{array}\]
\[\begin{array}{l}=\sec( {23}^\circ -\theta) - \sec( {23}^\circ - \theta) \\ \\ \end{array}\]
\[=0=R.H.S.\]
APPEARS IN
संबंधित प्रश्न
Show that:
(i) `2(cos^2 45º + tan^2 60º) – 6(sin^2 45º – tan^2 30º) = 6`
(ii) `2(cos^4 60º + sin^4 30º) – (tan^2 60º + cot^2 45º) + 3 sec^2 30º = 1/4`
Evaluate the following :
(sin 72° + cos 18°) (sin 72° − cos 18°)
Evaluate: `4(sin^2 30 + cos^4 60^@) - 2/3 3[(sqrt(3/2))^2 . [1/sqrt2]^2] + 1/4 (sqrt3)^2`
Evaluate: `(2sin 68)/cos 22 - (2 cot 15^@)/(5 tan 75^@) - (8 tan 45^@ tan 20^@ tan 40^@ tan 50^@ tan 70^@)/5`
If sin x = cos y, then x + y = 45° ; write true of false
prove that:
cos (2 x 30°) = `(1 – tan^2 30°)/(1+tan^2 30°)`
Prove that:
4 (sin4 30° + cos4 60°) -3 (cos2 45° - sin2 90°) = 2
If A = 30°;
show that:
cos 2A = cos4 A - sin4 A
If sin(A + B) = 1 and cos(A – B)= `sqrt(3)/2`, 0° < A + B ≤ 90° and A > B, then find the measures of angles A and B.
`(2/3 sin 0^circ - 4/5 cos 0^circ)` is equal to ______.
