Answer: [tex]\huge\colorbox{orange}{ANSWER}[/tex] Now, (cosecθ−sinθ)(secθ−cosθ) =(sinθ1−sinθ)(cosθ1−cosθ) =(sinθ1−sin2θ)(cosθ1−cos2θ) =sinθcosθcos2θsin2θ=sinθcosθ (1) Next, consider tanθ+cotθ1 =cosθsinθ+sinθcosθ1 =(sinθcosθsin2θ+cos2θ)1 =sinθcosθ (2) From (1) and (2), we get (cosecθ−sinθ)(secθ−cosθ)=tanθ+cotθ Reply
Here I am using A instead of theta. We know the trigonometric identities , 1 ) sin² A + cos² A = 1 2 ) sec² A = 1 + tan² A 3 ) cosec² A = 1 + cot² A And 4 ) sinA cosecA = 1 5 ) cosA secA = 1 Now , LHS = (sinA+cosecA)² + (cosA+secA)² = sin²A+cosec² A+2sinAcosecA+ cos² A + sec² A + 2cosAsecA = (Sin²A + cos² A ) + 2 + 2 + cosec² A + Sec² A = 1 + 2 + 2 + ( 1 + cot² A )+( 1 + tan²A ) = 7 + cot² A + tan² A = RHS I hope this helps you. Reply
Answer:
[tex]\huge\colorbox{orange}{ANSWER}[/tex]
Now,
(cosecθ−sinθ)(secθ−cosθ)
=(sinθ1−sinθ)(cosθ1−cosθ)
=(sinθ1−sin2θ)(cosθ1−cos2θ)
=sinθcosθcos2θsin2θ=sinθcosθ (1)
Next, consider tanθ+cotθ1
=cosθsinθ+sinθcosθ1
=(sinθcosθsin2θ+cos2θ)1
=sinθcosθ (2)
From (1) and (2), we get
(cosecθ−sinθ)(secθ−cosθ)=tanθ+cotθ
Here I am using A instead of theta.
We know the trigonometric
identities ,
1 ) sin² A + cos² A = 1
2 ) sec² A = 1 + tan² A
3 ) cosec² A = 1 + cot² A
And
4 ) sinA cosecA = 1
5 ) cosA secA = 1
Now ,
LHS = (sinA+cosecA)² + (cosA+secA)²
= sin²A+cosec² A+2sinAcosecA+
cos² A + sec² A + 2cosAsecA
= (Sin²A + cos² A ) + 2 + 2 + cosec² A
+ Sec² A
= 1 + 2 + 2 + ( 1 + cot² A )+( 1 + tan²A )
= 7 + cot² A + tan² A
= RHS
I hope this helps you.