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Question
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
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Solution
LHS = `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ)`
= `(tan θ)/(tan θ) + ( cosec θ. cos θ)/(cosθ. cosec θ)`
= 1 + 1 = 2
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
Prove that sec2θ − cos2θ = tan2θ + sin2θ
