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प्रश्न
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
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उत्तर
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80)`
`= (sin 35^@ . cos (90^@ - 35^@) + cos 35^@. sin (90^@ - 35^@))/(cosec^2(90^@ - 80^@) - tan^2 80^@`)
`= (sin 35^@ . sin 35^@ + cos 35^@ . cos 35^@) /(sec^2 80^@ - tan^2 80^@)`
`= (sin^2 35^@ + cos^2 35^@)/(sec^2 80^@ - tan^2 80^@) = 1/1 = 1`
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संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
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.
