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Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]?
Most Upvoted Answer
Prove the identity, where the angles involved are acute angles for whi...
LHS:tanθ1−cotθ+cotθ1−tanθLHS:tanθ1-cotθ+cotθ1-tanθ
=tanθ1−cosθsinθ+cotθ1−sinθcosθ=tanθ1-cosθsinθ+cotθ1-sinθcosθ
=sinθcosθsinθ−cosθsinθ+cosθsinθcosθ−sinθcosθ=sinθcosθsinθ-cosθsinθ+cosθsinθcosθ-sinθcosθ
=sinθcosθ×sinθsinθ−cosθ+cosθsinθ×cosθcosθ−sinθ=sinθcosθ×sinθsinθ-cosθ+cosθsinθ×cosθcosθ-sinθ
=sin2θcosθ(sinθ−cosθ)+cos2θsinθ(cosθ−sinθ)=sin2θcosθ(sinθ-cosθ)+cos2θsinθ(cosθ-sinθ)
=sin2θcosθ(sinθ−cosθ)+cos2θsinθ(−sinθ+cosθ)=sin2θcosθ(sinθ-cosθ)+cos2θsinθ(-sinθ+cosθ)
=sin2θcosθ(sinθ−cosθ)−cos2θsinθ(sinθ−cosθ)=sin2θcosθ(sinθ-cosθ)-cos2θsinθ(sinθ-cosθ)
=1sinθ−cosθ(sin2θcosθ−cos2θsinθ)=1sinθ-cosθ(sin2θcosθ-cos2θsinθ)
=1sinθ−cosθ(sin3θ−cos3θsinθ.cosθ)=1sinθ-cosθ(sin3θ-cos3θsinθ.cosθ)
=1sinθ−cosθ×(sinθ−cosθ)(sin2θ+cos2θ+sinθ.cosθ)sinθ.cosθ=1sinθ-cosθ×(sinθ-cosθ)(sin2θ+cos2θ+sinθ.cosθ)sinθ.cosθ
=sin2θ+cos2θ+sinθ.cosθsinθ.cosθ=sin2θ+cos2θ+sinθ.cosθsinθ.cosθ
=1+sinθ.cosθsinθ.cosθ=1+sinθ.cosθsinθ.cosθ
=1sinθ.cosθ+sinθ.cosθsinθ.cosθ=1sinθ.cosθ+sinθ.cosθsinθ.cosθ
=1sinθ.cosθ+1=1sinθ.cosθ+1
=secθ.cosecθ+1=secθ.cosecθ+1
=1+sec&θ.cosecθ=RHS=1+sec&θ.cosecθ=RHSproved
=tanθ1−cosθsinθ+cotθ1−sinθcosθ=tanθ1-cosθsinθ+cotθ1-sinθcosθ
=sinθcosθsinθ−cosθsinθ+cosθsinθcosθ−sinθcosθ=sinθcosθsinθ-cosθsinθ+cosθsinθcosθ-sinθcosθ
=sinθcosθ×sinθsinθ−cosθ+cosθsinθ×cosθcosθ−sinθ=sinθcosθ×sinθsinθ-cosθ+cosθsinθ×cosθcosθ-sinθ
=sin2θcosθ(sinθ−cosθ)+cos2θsinθ(cosθ−sinθ)=sin2θcosθ(sinθ-cosθ)+cos2θsinθ(cosθ-sinθ)
=sin2θcosθ(sinθ−cosθ)+cos2θsinθ(−sinθ+cosθ)=sin2θcosθ(sinθ-cosθ)+cos2θsinθ(-sinθ+cosθ)
=sin2θcosθ(sinθ−cosθ)−cos2θsinθ(sinθ−cosθ)=sin2θcosθ(sinθ-cosθ)-cos2θsinθ(sinθ-cosθ)
=1sinθ−cosθ(sin2θcosθ−cos2θsinθ)=1sinθ-cosθ(sin2θcosθ-cos2θsinθ)
=1sinθ−cosθ(sin3θ−cos3θsinθ.cosθ)=1sinθ-cosθ(sin3θ-cos3θsinθ.cosθ)
=1sinθ−cosθ×(sinθ−cosθ)(sin2θ+cos2θ+sinθ.cosθ)sinθ.cosθ=1sinθ-cosθ×(sinθ-cosθ)(sin2θ+cos2θ+sinθ.cosθ)sinθ.cosθ
=sin2θ+cos2θ+sinθ.cosθsinθ.cosθ=sin2θ+cos2θ+sinθ.cosθsinθ.cosθ
=1+sinθ.cosθsinθ.cosθ=1+sinθ.cosθsinθ.cosθ
=1sinθ.cosθ+sinθ.cosθsinθ.cosθ=1sinθ.cosθ+sinθ.cosθsinθ.cosθ
=1sinθ.cosθ+1=1sinθ.cosθ+1
=secθ.cosecθ+1=secθ.cosecθ+1
=1+sec&θ.cosecθ=RHS=1+sec&θ.cosecθ=RHSproved
Community Answer
Prove the identity, where the angles involved are acute angles for whi...
Proof:
To prove the given identity, we will start by expressing the expression in terms of sin and cos.
Let's first rewrite the expression on the left-hand side (LHS) of the equation:
LHS = (tanθ / (1 - cotθ)) * (cotθ / (1 - tanθ))
Now, we can rewrite tanθ and cotθ in terms of sinθ and cosθ:
tanθ = sinθ / cosθ
cotθ = cosθ / sinθ
Substituting these values into the expression, we get:
LHS = [(sinθ / cosθ) / (1 - (cosθ / sinθ))] * [(cosθ / sinθ) / (1 - (sinθ / cosθ))]
Simplifying further, we can rewrite this as:
LHS = [(sinθ / cosθ) / ((sinθ - cosθ) / sinθ)] * [(cosθ / sinθ) / ((cosθ - sinθ) / cosθ)]
Now, let's simplify each fraction separately:
LHS = [(sinθ / cosθ) * (sinθ / (sinθ - cosθ))] * [(cosθ / sinθ) * (cosθ / (cosθ - sinθ))]
LHS = [(sinθ * sinθ) / (cosθ * (sinθ - cosθ))] * [(cosθ * cosθ) / (sinθ * (cosθ - sinθ))]
LHS = [(sinθ * sinθ * cosθ * cosθ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
Notice that sinθ * cosθ = sinθ * (-sinθ) = -sin²θ
LHS = [(-sin²θ * cos²θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
Using the identity: sin²θ + cos²θ = 1, we can substitute sin²θ with (1 - cos²θ):
LHS = [(-(1 - cos²θ) * cos²θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
LHS = [(-cos²θ + cos⁴θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
Simplifying the numerator further:
LHS = (-cos²θ + cos⁴θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))
LHS = (cos⁴θ - cos²θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))
Using the identity: cos²θ = 1 - sin²θ, we can substitute cos²θ with (1 - sin²θ):
LHS = [(1 - sin²θ) * (1 - sin²θ)] / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))
LHS = [1 - 2sin²θ + sin⁴θ] / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sin
To prove the given identity, we will start by expressing the expression in terms of sin and cos.
Let's first rewrite the expression on the left-hand side (LHS) of the equation:
LHS = (tanθ / (1 - cotθ)) * (cotθ / (1 - tanθ))
Now, we can rewrite tanθ and cotθ in terms of sinθ and cosθ:
tanθ = sinθ / cosθ
cotθ = cosθ / sinθ
Substituting these values into the expression, we get:
LHS = [(sinθ / cosθ) / (1 - (cosθ / sinθ))] * [(cosθ / sinθ) / (1 - (sinθ / cosθ))]
Simplifying further, we can rewrite this as:
LHS = [(sinθ / cosθ) / ((sinθ - cosθ) / sinθ)] * [(cosθ / sinθ) / ((cosθ - sinθ) / cosθ)]
Now, let's simplify each fraction separately:
LHS = [(sinθ / cosθ) * (sinθ / (sinθ - cosθ))] * [(cosθ / sinθ) * (cosθ / (cosθ - sinθ))]
LHS = [(sinθ * sinθ) / (cosθ * (sinθ - cosθ))] * [(cosθ * cosθ) / (sinθ * (cosθ - sinθ))]
LHS = [(sinθ * sinθ * cosθ * cosθ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
Notice that sinθ * cosθ = sinθ * (-sinθ) = -sin²θ
LHS = [(-sin²θ * cos²θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
Using the identity: sin²θ + cos²θ = 1, we can substitute sin²θ with (1 - cos²θ):
LHS = [(-(1 - cos²θ) * cos²θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
LHS = [(-cos²θ + cos⁴θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))]
Simplifying the numerator further:
LHS = (-cos²θ + cos⁴θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))
LHS = (cos⁴θ - cos²θ) / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))
Using the identity: cos²θ = 1 - sin²θ, we can substitute cos²θ with (1 - sin²θ):
LHS = [(1 - sin²θ) * (1 - sin²θ)] / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sinθ))
LHS = [1 - 2sin²θ + sin⁴θ] / (cosθ * sinθ * (sinθ - cosθ) * (cosθ - sin
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Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]?
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Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]? 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 Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]?.
Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]? 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 Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]?.
Solutions for Prove the identity, where the angles involved are acute angles for which the following expression is defined:- Tan theta/1-cot theta + cot theta/1-tan theta=1 sec theta cosec theta [Hint: Write the expression in term of sin theta and cos theta]? in English & in Hindi are available as part of our courses for Class 10.
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