Question

please show me the steps​

Answer 1

EXPLANATION.

• step =1

we can firstly simplify this term of L. H. S

• step = 2

arrange the term of L. H. S

• step = 3

put the value of this term

Therefore,

L. H. S = R. H. S

formula used

$1) = (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab$

$2) = \csc( \theta) = \frac{1}{ \sin( \theta) }$

$3) = \sec( \theta) = \frac{1}{ \cos( \theta) }$

$4) = \sin {}^{2} ( \theta) + \cos {}^{2} ( \theta) = 1$

$5) = \csc {}^{2} ( \theta) = 1 + \cot {}^{2} ( \theta)$

$6) = \sec {}^{2} ( \theta) = 1 + \tan {}^{2} ( \theta)$

note = see image for the solution of this question.

Answer 2

$\orange{\bold{\underline{\underline{Given:-}}}}$Question :

(sin A +cosec A )²+(cos A+sec A)² + (cos A+sec A)² = 7 + tan²A + cot²A

$\blue{\bold{\underline{\underline{To \: find:-}}}}$

• LHS = RHS

FORMULAE Used:

• 1+tan²A=sec²A

• 1+cot²A=cosec²A

• sin²A+cos²A=1

• cosec A = 1 / sin A

• sec A = 1 / cos A

Lets Solve with LHS.

LHS =(sinA+cosecA)²+(cosA+secA)²

=sin²A+cosec²A+2sinAcosecA+cos²A+sec²A+2cosAsecA

=sin²A+cos²A+cosec²A+sec²A+2sinA×1/sinA+2cosA×1/cosA

=1+cosec²A+sec²A+2+2

=5+(1+cot²A)+(1+tan²A)

=7+tan²A+cot²A

Hence, LHS = RHS

$\bf\blue{Hope\ it\ helps.}$

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