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प्रश्न
If θ is an acute angle of a right angled triangle, then which of the following equation is not true?
पर्याय
sin θ cot θ = cos θ
cos θ tan θ = sin θ
cosec2 θ – cot2 θ = 1
tan2 θ – sec2 θ = 1
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उत्तर
tan2 θ – sec2 θ = 1
Explanation:
tan2 θ – sec2 θ = 1 is not true
∵ sec2 θ = 1 + tan2 θ
or sec2 θ – tan2 θ = 1
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संबंधित प्रश्न
In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.
`sec theta = 13/5`
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If `tan theta = 24/7`, find that sin 𝜃 + cos 𝜃
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Evaluate the following
cos 60° cos 45° - sin 60° ∙ sin 45°
Evaluate the following
`sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°`
Evaluate the Following
`cot^2 30^@ - 2 cos^2 60^circ- 3/4 sec^2 45^@ - 4 sec^2 30^@`
Find the value of x in the following :
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Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.
Proof: L.H.S. = sec θ + tan θ
= `1/square + square/square`
= `square/square` ......`(∵ sec θ = 1/square, tan θ = square/square)`
= `((1 + sin θ) square)/(cos θ square)` ......[Multiplying `square` with the numerator and denominator]
= `(1^2 - square)/(cos θ square)`
= `square/(cos θ square)`
= `cos θ/(1 - sin θ)` = R.H.S.
∴ L.H.S. = R.H.S.
∴ sec θ + tan θ = `cos θ/(1 - sin θ)`
