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рдкреНрд░рд╢реНрди
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
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рдЙрддреНрддрд░
LHS= `(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta)`
=` ((1+ cos theta )- (1-cos^2 theta))/(sin theta(1+ cos theta))`
=`(cos theta + cos^2 theta)/( sin theta ( 1+ cos theta))`
=`(cos theta ( 1+ cos theta ))/ ( sin theta ( 1+ cos theta))`
=`cos theta/ sin theta`
= cot ЁЭЬГ
= RHS
Hence, L.H.S. = R.H.S.
APPEARS IN
рд╕рдВрдмрдВрдзрд┐рдд рдкреНрд░рд╢реНрди
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.
Write the value of tan10° tan 20° tan 70° tan 80° .
Write the value of cos1° cos 2°........cos180° .
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
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`
