Advertisements
Advertisements
प्रश्न
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
उत्तर
LHS = `cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°)`
= `cos^2 26° + cos (90° - 26°) sin 26° + (tan 36°)/(cot (90° - 54°)`
= `cos^2 26° + sin 26°. sin 26° + (tan 36°)/(tan 36°)`
= cos2 26° + sin2 26 + 1 ....( cos2 θ + sin2 θ = 1)
= 1 + 1 = 2
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
If 2 cos θ + sin θ = `1(θ ≠ π/2)`, then 7 cos θ + 6 sin θ is equal to ______.
(1 – cos2 A) is equal to ______.
