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प्रश्न
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
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उत्तर
LHS = `cosA/(1 - tanA) + sinA/(1 - cotA)`
= `cosA/(1-sinA/cosA) + sinA/(1 - cosA/sinA) = cosA/((cosA -sinA)/cosA) + sinA/((sinA - cosA)/sinA)`
= `cos^2A/(cosA - sinA) + sin^2A/(sinA - cosA) = (cos^2A - sin^2A)/((cosA - sinA))`
`((cosA - sinA)(cosA + sinA))/(cosA - sinA)`
= sinA + cosA = RHS
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संबंधित प्रश्न
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(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
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`(1 + cot^2 theta ) sin^2 theta =1`
`(sec^2 theta-1) cot ^2 theta=1`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
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`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
