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प्रश्न
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
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उत्तर
`cos^2 26^@ + cos 64^@ sin 26^@ + tan 36^@/cot 54^@`
`= cos^2 26^@ + cos(90^@ - 26^@) sin 26^@ + tan 36^@/(cot(90^@ - 36^@))`
`= cos^2 26^@ + sin 26^@.sin26^@ + tan36^@/tan36^@` `[∵ cos(90^@ - theta) = sin theta, cot(90^@ - theta) = tan theta]`
`= cos^2 26^@ + sin^2 26^@ + 1`
`= 1 + 1 [∵ cos^2 theta + sin^2 theta = 1]`
= 2
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संबंधित प्रश्न
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
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`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
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`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
