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рдкреНрд░рд╢реНрди
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
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рдЙрддреНрддрд░
ЁЭР┐ЁЭР╗ЁЭСЖ = `(tan theta)/(1+tan^2 theta )^2 +( cot theta )/(1+cot^2 theta)^2`
=`tan theta/ ((sec^2 theta)^2) + cot theta/((cosec^2 theta) ^2)`
=`tan theta / sec^4 theta + cottheta/(cosec^4 theta)`
=`sin theta/cos theta xx cos^4 theta + cos theta/sin theta xx sin ^4 theta`
=` sin theta cos ^3 theta + cos theta sin ^3 theta`
=`sin theta cos theta ( cos^2 theta + sin ^2 theta)`
=`sin theta cos theta`
= RHS
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рд╕рдВрдмрдВрдзрд┐рдд рдкреНрд░рд╢реНрди
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
If sin θ = `11/61`, find the values of cos θ using trigonometric identity.
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
1 + cot2θ = ?
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ.
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
