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Show that cos2(45∘+θ)+cos2(45∘-θ)tan(60∘+θ)tan(30∘-θ) = 1 - Mathematics

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प्रश्न

Show that `(cos^2(45^circ + theta) + cos^2(45^circ - theta))/(tan(60^circ + theta) tan(30^circ - theta))` = 1

योग

उत्तर

L.H.S =  `(cos^2(45^circ + theta) + cos^2(45^circ - theta))/(tan(60^circ + theta) * tan(30^circ - theta))` 

= `(cos^2(45^circ + theta) + [sin{90^circ - (45^circ - theta)}]^2)/(tan(60^circ + theta) * cot{90^circ - (30^circ - theta)})`   ...[∵ sin(90° – θ) = cos θ and cot(90° – θ) = tan θ]

= `(cos^2(45^circ + theta) + sin^2(45^circ + theta))/(tan(60^circ + theta) * cot(60^circ + theta))`  ...[∵ sin2θ + cos2θ = 1]

= `1/(tan(60^circ + theta) * 1/(tan(60^circ + theta))`  ...`[∵ cot θ = 1/tanθ]`

= 1

= R.H.S

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Introduction To Trigonometry and Its Applications - Exercise 8.3 [पृष्ठ ९५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
अध्याय 8 Introduction To Trigonometry and Its Applications
Exercise 8.3 | Q 13 | पृष्ठ ९५

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