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Question
जर tan θ + cot θ = 2, तर tan2θ + cot2θ = ?
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Solution
tan θ + cot θ = 2 ....[दिलेले]
∴ (tan θ + cot θ)2 = 4 .....[दोन्ही बाजूंचा वर्ग करून]
∴ tan2θ + 2tan θ.cot θ + cot2θ = 4 ......[∵ (a + b)2 = a2 + 2ab + b2]
∴ tan2θ + 2(1) + cot2θ = 4 ......[∵ tan θ ⋅ cot θ = 1]
∴ tan2θ + cot2θ = 4 – 2
∴ tan2θ + cot2θ = 2
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कृती: डावी बाजू = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= उजवी बाजू
`(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ हे सिद्ध करा.
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A हे सिद्ध करा.
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
जर sin θ + cos θ = `sqrt(3)`, तर tan θ + cot θ = 1 हे दाखवा.
(1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B हे सिद्ध करा.
जर `1/sin^2θ - 1/cos^2θ-1/tan^2θ-1/cot^2θ-1/sec^2θ-1/("cosec"^2θ) = -3`, तर θ ची किमत काढा.
