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Question
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
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Solution
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.
RELATED QUESTIONS
`sin^2 theta + 1/((1+tan^2 theta))=1`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
If x= a sec `theta + b tan theta and y = a tan theta + b sec theta ,"prove that" (x^2 - y^2 )=(a^2 -b^2)`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Find the value of sin 30° + cos 60°.
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
