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(sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²?
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(sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . c...
Let angle theta be represented by A.
(sinA + Sec A)² + (cos A + cosec A)²
= [ (SinA CosA + 1)²/CosA ]² + [ (CosA SinA + 1)² / Sin²A ]
= (1+ SinA CosA)² * [1 / Cos²A + 1/sin²A ]
= (1 + SinA CosA)² * [ Cos²A + Sin²A]/ [cos²A * Sin²A ]
= ( 1 + SinA CosA)²/ (cosA * SinA)²
= [ 1/cosA * 1/sinA + 1 ] ²
= [ Sec A Cosec A + 1 ]²
(sinA + Sec A)² + (cos A + cosec A)²
= [ (SinA CosA + 1)²/CosA ]² + [ (CosA SinA + 1)² / Sin²A ]
= (1+ SinA CosA)² * [1 / Cos²A + 1/sin²A ]
= (1 + SinA CosA)² * [ Cos²A + Sin²A]/ [cos²A * Sin²A ]
= ( 1 + SinA CosA)²/ (cosA * SinA)²
= [ 1/cosA * 1/sinA + 1 ] ²
= [ Sec A Cosec A + 1 ]²
Community Answer
(sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . c...
Proving (sin theta sec theta)² (cos theta cosec theta)² = (1 sec theta cosec theta)²
Step 1: Simplification of LHS
Let's expand the left-hand side of the equation:
(sin theta sec theta)² (cos theta cosec theta)² = (sin² theta / cos² theta) (cos² theta / sin² theta)
Using the identity sin² theta + cos² theta = 1 and dividing both numerator and denominator of each term by cos² theta and sin² theta, respectively, we get:
(sin² theta / cos² theta) (cos² theta / sin² theta) = (sin² theta + cos² theta) / (cos² theta × sin² theta)
Now, using the identity sec theta = 1 / cos theta and cosec theta = 1 / sin theta, we can simplify the above expression as:
(sin² theta + cos² theta) / (cos² theta × sin² theta) = 1 / (cos² theta × sin² theta) = (1 / cos theta)² × (1 / sin theta)²
Therefore, the left-hand side of the equation simplifies to:
(sin theta sec theta)² (cos theta cosec theta)² = (1 / cos theta)² × (1 / sin theta)²
Step 2: Simplification of RHS
Now, let's simplify the right-hand side of the equation:
(1 sec theta cosec theta)² = (1 / cos theta × 1 / sin theta)² = (1 / (cos theta × sin theta))²
Using the identity sin 2θ = 2 sin θ cos θ and cos 2θ = cos² θ - sin² θ, we can write:
sin 2θ = 2 sin θ cos θ = 2 (sin θ / cos θ) × sin θ = 2 sin² θ / cos θ
cos 2θ = cos² θ - sin² θ = (cos θ / cos θ)² - (sin θ / cos θ)² = (1 - sin² θ / cos² θ)
Now, using the above identities, we can write:
(1 / (cos theta × sin theta))² = (cos² θ + sin² θ)² / (4 cos² θ sin² θ) = (1 + sin² θ / cos² θ)² / 4
Using the identity sec theta = 1 / cos theta and cosec theta = 1 / sin theta, we can further simplify the above expression as:
(1 + sin² θ / cos² θ)² / 4 = (1 + sec² θ)² / 4 = (1 / cos² θ)² / (1 / sin² θ)² = (1 /
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(sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²?
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(sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about (sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²?.
(sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about (sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (sin theta sec theta)² ( cos theta cosec theta)² =(1 sec theta . cosec theta)²?.
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