Question

sin²Q tanQ + cos²Q cotQ + 2sinQ cosQ = tanQ + cotQ



pls give answer i will mark brainlist​

Answer 1

Step-by-step explanation:

Given :-

sin²Q tanQ + cos²Q cotQ + 2sinQ cosQ

To find :-

Prove that :

sin²Q tanQ + cos²Q cotQ + 2sinQ cosQ = tanQ + cotQ

Solution:-

On taking LHS :-

sin²Q tanQ + cos²Q cotQ + 2sinQ cosQ

=> sin²Q (sinQ/cosQ) + cos²Q (cosQ/sinQ)

+ 2sinQ cosQ

Since Tan A = Sin A/ Cos A and

Cot A = Cos A/ Sin A

=>(sin³Q/cosQ)+(cos³Q/sinQ)+2sinQcosQ

=>(sin³Q sinQ+ cos³QcosQ+2sinQ cosQ . sinQ cosQ)/(sin Q cos Q)

=>(sin⁴Q+cos⁴Q+2sin²Qcos²Q)/(sinQcosQ)

=>[(sin²Q)² + 2sin²Q cos²Q + (cos²Q)²]/(sinQcosQ)

=> [sin²Q+cos²Q]² /(sinQcosQ)

Since (a+b)² = a²+2ab+b²

Where , a = sin²Q and b = cos²Q

We know that

Sin² A + Cos² A = 1

=> (1)²/(sinQcosQ)

=>( 1/sin Q )(1/cos Q)

=> LHS = Cosec Q . sec Q -----------(1)

On taking RHS :-

Tan Q + cot Q

=> (sin Q /Cos Q ) + ( cos Q / sin Q)

=> (sinQsinQ + cosQcosQ ) /( sin Q.cos Q)

=>( sin² Q + cos² Q)/(sin Q cos Q)

=> 1/(sin Q . cos Q)

Since Sin² A + Cos² A = 1

=>( 1/sin Q )(1/cos Q)

=> RHS = Cosec Q . sec Q -----------(2)

From (1) &(2)

LHS = RHS

Hence, Proved.

Answer:-

sin²Q tanQ + cos²Q cotQ + 2sinQ cosQ

= tanQ + cotQ

Used formulae:-

• Tan A = Sin A / Cos A
• Cot A = Cos A / Sin A
• 1 / Sin A = Cosec A
• 1 / Cos A = Sec A
• Sin² A + Cos² A = 1
• (a+b)² = a²+2ab+b²