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प्रश्न
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
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उत्तर
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.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
If 5 `tan theta = 4,"write the value of" ((cos theta - sintheta))/(( cos theta + sin theta))`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
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
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
